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CONTINUITY IN CONCRETE BUILDINQ FRflflES Practical Analysis for Vertical Load and Wind Pressure PORTLAND CEMENT ASSOCIATION ' CONTINUITY IN CONCRETE BUILDINQ FRdnES ;• Tribune Building Montevideo, S* A* Published by PORTLAND CEMENT ASSOCIATION Concrete for Permanenc e 33 WEST GRAND AVENUE . CHICAGO, ILLINOIS [BLANK PAGE] CCA INWH NATIQNAJ PREFACE The common practice of designing beams in building frames by use of moment coefficients and ignoring moments in columns is not always safe and adequate. It may result in distress in many structures and over-design in others. It disregards many important elements including concentrated loads, ratio of live load to dead load, ratio of span lengths in adjacent bays, ratio of column stiffness to beam stiffness, and ratio of column width to span length. The effect of all these elements upon design cannot be incor- porated in three simple beam moments such as l /i Q wl 2 , l /nwl 2 and l /uwl 2 . One reason why the moment coefficients are still used is that making a proper analysis has been a complex and lengthy task. This booklet is prepared 10 facilitate the analysis and to discuss quite fully the elements that go into the making of a well-balanced design. More specifically, the aim is to give constructive suggestions for improved practice and to present analytical procedures that are practical, i.e., quick, easy, convenient and yet sufficiently accurate for office design practice. Considering the reduction in labor by the use of the proposed procedures, which are approximate, the accuracy of the results is surprisingly good. Even procedures that are mathematically exact do not yield exact results because they rest on assumptions that are approximate. In fact, better accuracy may be obtained from "approximate" analysis based on good assumptions than from "exact" analysis based on poor assumptions. Good judgment and thorough understanding of the assumptions under- lying the analysis are essential For this reason, the presentation goes beyond the mere statement of analytical procedures, which are simple and occupy limited space only. Included are also: underlying assumptions, dis- cussion and special studies, most of the latter in form of numerical prob- lems. Deriv ii ions that are not essential to the application of the procedures have been deferred to the appendixes. The problems are recommended for careful study because they make the render familiar with the simplicity of the procedure, the accuracy obtainable and the effect of varying assumptions. Procedures are given for both vertical load and wind pressure. The dis- cussion of wind pressure applies to earthquake analysis also, if earthquake shocks are treated in the customary manner as static horizontal loading. The analysis presented for vertical load is essentially Professor Cross' method of moment distribution applied— with slight modifications — to two joints only. For very irregular cases, such as offset columns supported on g transfer girders, the procedure may have to be extended by using moment distribution applied to more than two joints. Also for wind pressure, the highly irregular cases need special study beyond that given in this text. 3 O P December. 1935 L\ [BLANK PAGE] CCA IN T F NNATtONAl CONTENTS NOTATION VERTICAL LOAD General Procedure of Analysis 1. Introduction ......... 2. General procedure of analysis . 3. Selection of Q-values when cross-sections are not known 4. Examples of analysis ........ 5. Moments in beams in roofs and bottom floors 6. Shear in beams ........ 7. Column moments and beam moments at wall columns Special Procedure of Analysis 8. Assumptions regarding restraint at far ends of columns 9. Correction for rotation at far ends of columns 10. Summary ) Discussion of Various Effects 11. Effect of width of columns ...... 12. Effect of change in moment of inertia .... 13. Effect of distant loads ....... It. Effect of haunching ........ Page 5 6 7 9 15 15 16 19 20 23 24 26 29 29 WIND PRESSURE 15. Introduction ...... # ... 31 16. Illustrative problem ........ 32 17. Eccentric wind pressure ....... 35 18. Accuracy of approximate analysis ...... 37 19. Moment deflection and the ideal layout .... 39 20. Illustrative problem^ vertical load and wind pressure combined 40 APPENDIX A: Derivation of Formulas— Vertical Load 21. Angle changes determined by moment area principle . . 45 22. Angle changes due to end moments ..... 46 23. End moments required for fixity ...... 47 24. Moments as a function of end rotation and fixed end moments 48 25. Joint rotation a function of stiffness ..... 49 26. Derivation of formulas for joint coefficient, Q . . . 50 27. Conventions and use of signs ...... 51 28. Formulas for moments in columns ..... 53 APPENDIX B: Derivation of Formulas— Wind Pressure 29. Assumptions and distribution of wind pressure ... 53 BIBLIOGRAPHY NOTATION M = moment. Mab = end moment at joint A of member AB. M F = fixed end moment. U = unbalanced moment. W = total load on span. w = load per linear foot. P = concentrated load or axial load. I — span length. h = story height. J = moment of inertia. K = I /I or I/h. n = ratio of U K for column" to "K for beam" Q = a joint coefficient. e = eccentricity. d = depth of section. a = width of supporting member. b = width of section. A = area. c = diameter of circle. V = end shear. v = relative shear. = angle of rotation at end of member. E = modulus of elasticity. / = a coefficient expressing degree of restraint R = angle of joint translation. Continuity in Concrete Building Frames Vertical Load General Procedure of Analysis 1. Introduction A procedure is presented for determination of end moments in horizontal members of continuous building frames. It has certain characteristics in common with current analytical methods, and the underlying principles in the derivation — given in Appendix A — are the same as in, for example, the slope-deflection and the moment-distribution methods. Current ana- lytical methods are not superseded but merely adapted so that results with satisfactory degree of accuracy may be obtained by the least amount of work. In analysis of continuous frames, the concept of fixed end moments is especially useful; they are the moments required to fix the ends of members, or — as is sometimes said — to lock the joints in frames. Fixed end moments, M F , may be expressed in terms of the loading, W, and the span length, I; and M F for one member is independent of other members. Moment coeffi- cients for prismatic beams* with fixed ends and subject to nineteen types of loading are compiled in Fig. 1, in which the example in the lower right corner illustrates the use of the moment coefficients. While determining values of M F as, for example, in Fig. 2, joints C, A, B and D are assumed to be temporarily locked (prevented from rotat- ing). If the artificial restraint is removed, each joint will rotate under the influence of the unbalanced moment: the difference between the fixed end moments. The use of unbalanced moments — denoted as U — is convenient in frame analysis, and U may readily be determined from values of M F . When joints rotate, new moments are induced at the ends of the mem- bers, and the final moment is equal to the sum of M F and "induced moments." The problem to be solved by analysis of typical building frames is principally to determine the induced moments. The induced moments are usually small compared with M F . They may — in a sense — be considered as a correction to M F } and a high degree of accuracy in the determination of induced moments is not essential. These moments will be approximated in the procedure which follows, and the work involved in the frame analysis will thereby be greatly reduced. ♦Members will be assumed prismatic except in Section 14: Effect of Haunching. 5 Coefficients of W*l Symmetrical loading i y i i- 2: 3- 4: 5- e- 7- 8- 9: ZO- lh t (Jnsymmetrical loading W--{wl / 3 12'- k 10 _ J_ IS I w 7 W 7 3^T3 2 9 f3 '9& i . — 4 "j*V ~JT4 ~^T4 Each lodd:W 3 _±l±f±fZZLj a I 5 10 a la I* a(G-8a+3a z ) 12 h. Ji z W-wal 96 1-3 1 a H4-3a) n I - 12 15: afl-a) z a! l-al aH/-a) W=wl t m^ i r ii 16 23 740 mi iff i ,n. a(IO-IOa+3a l ) U ' 30 7 al l-al a 2 (S-3a) l*jk - 1 IV--ju/l w * l 9: 3fl0-/5a*6a*) LrtiMK A J 40 W -Jural 3 1 l-al 9H5-43) 16 fxamp/e : 4rr •5/ JV-- 2wal T 20 0tf<? IE I y*t -^4- /5 : ± T 3 /iW7 i ■l-2al^l-al W--w{l-2all n ■alA—l-Zal^ral a (3-7 a) Aa?oo*?o)?o* 4o,ooo k (i*800* 20)70 --16, 000 I* 7a -2a 2 if JM20.000) 2 46,000 Fixed End Moment at & ,1(1200*20)20-40,000 ,i (1*800*20)70-- 10,100 (iM(mOO0) 70 * 18,700 12 Totah 117,000 ft lb. Tot ah 69. 400 ft. lb. Fig. 1* Fixed End Moments — in Terms of Wl — for Prismatic Beams 2. General Procedure of Analysis The value of M F for prismatic members is a function of loads, W, and span lengths, I; but an induced moment depends also upon the moment of inertia, /, of the members in the frame. Let K denote the ratio of I /I for individual members, and n the ratio of K- value of a column to if-value *By permission of the authors, Fig. 1 is taken — with slight modifications — from Fig. 5, page 85, in 'Continuous Frames of Reinforced Concrete," by Cross and Morgan, published by John Wiley and Sons. 1932. 6 of a beam. Equations for n A and n B are given in Fig. 2. This figure illustrates the general procedure which is described in Steps (1) to (6): (1) Record in Fig. 2 the loading on the three adjacent spans, 1 C a, Iab and l BD . (2) From this loading, determine values of M F (using Fig. 1 ) and record M F as negative quantities in Fig. 2. (3) Compute the unbalanced moment at A, Ua, and at B, U B , as M F to the left minus M F to the right of each joint. Record U with proper sign. (4) Compute (or select, as explained in Section 3) values of n or Q (see Fig. 2 for formula for Q* applying to the case shown). (5) Record products of U times Q as shown and with proper signs as in Fig. 2. (6) Determ in e the final end moment as the algebraic sum of the three terms in the spaces marked X. The calculations in Steps (1), (2) and (3) are those that precede most methods of analysis. Steps (4), (5) and (6) comprise the analytical work characteristic of the procedure in this text and will be discussed further in subsequent sections. T I CA 4- l AB» h I A3 i so c < Wc-MIb Record the loading here * A A m *2Q**U* Ti AB *0m'U, B M M Q**u A -2Q 6 'U a 1 d£ Q = „ ' n CSee F/G.3) M £A D ML M 8A~ M 6D^'d 4n*-2 I 8 Negative moments give tension in top of beams. I A , I 3 and h are average raiues for columns above and below joints A and 5. Note that LI • moment to the left of joint minus moment to the right of joint. Fig. 2 5. Selection of Q-Values When Cross- Sections Are Not Known The relative values of K — denoted as n — must be chosen before a frame can be analyzed. How to select n, and the joint coefficient Q } is a question of the greatest importance to the designer. In reviewing a frame already designed, the procedure is simply to com- pute K, n, Q and then proceed with the analysis. This is the type of problem *See Appendix A for derivation. usually treated in papers and textbooks. When only loads, spans and story heights are known, the values of Q must be selected. The reader who is familiar with Professor Cross' moment distribution method* will recognize in the procedure in Fig. 2 the same characteristics as in that method: fixed end moments {M F ) t unbalanced moments (U), carry-over factors, and distribution factors (Q). A number of cases were 1 studied in most of which a Q-factor of 4n + 2 gave better results for frames with two joints than the corresponding factor, , used for regular 1 4n+ 2 0.5 0.4 0.3 moment distribution in frames with many joints. The equations Q = 1 proposed for joints in floors and Q —- — ^r~z proposed in Section 5 for joints in roofs are derived empirically consistent with the expressions for dis- tribution factor known from the moment distribution method. A refine- ment of giving different Q-formulas for interior and exterior joints is deemed unwarranted. According to the equation in Fig. 2, Q equals Yi when n = (no column restraint) and decreases to zero with increasing column stiff- ness. The variation of Q between the limits of Y and zero is plotted in Fig. 3. When n is greater than about 4, the Q- values are small and vary only slightly. In this range, which includes the lower stories of tall buildings, Q = V20 is a good average value . Part of the Q-diagram near the ordinate axis (n = in Fig. 3) corre- sponds to slightly restrained con- tinuous girders. For ordinary building frames, n need seldom be taken smaller than one-half,** and the value of Q = YL may be selected as typical for frames with slender columns having n-values smaller than unit> When n is between 1 and 4, the value of Q = }/$ may be considered typ- ical. This is the range between the very slender and the very stubby columns. If cross-sections arc not known in ordinary building frames, the following suggestions may be used as a guide in selecting values of Q: 0.2 0.1- ( No CO) 'umn ' res fraint 1 1 \ Diagran 7 IS V~4k 1 1+2 \ 5»_ % <^* n- valu es 4 Fig. 3 £ to Values of n Below 1 Between 1 and 4 Above 4 Column Classification Slender Medium Stubby Values of Q M H Vm ee References 10 and IK **See CoDclusionB in Section 4. <'! 4* Examples of Analy Problem 1. The frame in Fig. 4(a) has equal spans 20 ft. long and equal story heights; dead load equals 1000 lb. and live load 1200 lb. per lin. ft. Determine the maximum center moments in span AB and the corresponding end moment, M A b- The analysis will be carried out according to steps (1) to (6) on page 7 and arranged as in Fig. 2. Maximum center moment in AB is produced by placing live load in AB and also in alternate spans as in Fig. 4(a).* This determines the loadings and the fixed end moments, M F . The beam and column dimensions are not known. Suppose the columns will be very slender and select Q = % (see Section 3). From the loading and span given, compute the fixed end moments at A and B by using the coefficient in case 5 in Fig. 1, which gives M F = - l / 12 W X I = - l / n wV. All fixed end moments are negative and are recorded in 1000 ft. lb. in the computation form below. The unbalanced moment, U, defined in Section 2 as the M F to the left minus the M F to the right, equals (-33) - ( (-73) - ( 73) = +40 at A, and 33) = -40 at B. The corrections to the fixed end moment due to the unbalanced moment of +40 at A equals +2 X QU = +2 X % (+40) = +20 at A, and -1 X QU = -1 X M (+40) = -10 at B. The unbalanced moment of —40 at B gives the following corrections: +1 X QU = +1 X M (-40) = -10 at A, and -2 X QU = -2 X M (-40) = +20 at B. The final moments, M AB and M BA , equal the algebraic sum of M F and the two corrections. n = H n = H w=\000, 1=20 1 A 10=2200, f=20 1 B w='\0Q0,l=20 \S 1/ -VaX1 000X20-' -33 -73 -Vi2X2200X20 2 ■ - -Vi2X2200X20 2 I -73 -33 -VijX1000X20 2 (-33)-(-73) +40 +20 +2XKX(+40) -KXC+40) -10 -10 + KXC-40) --2XKX(-40) +20 -40 <-73)-(-33> -63 M AB M BA -63 Maximum positive moment in span AB is + H X 2200 X 20 2 - 63,000 = +110,000 - In this case, due to symmetry, the moment been simplified as follows: 63,000 = +47,000 ft.lb. computations might have ♦Maximum center moment requires (1) full live load on AB and (2) minimum restraint at joints A and B; minimum restraint from beams adjacent to A and B is obtained when they carry no live load; and minimum restraint from columns adjacent to joints A and B requires load on alternate spans as shown on floor above and below floor AB in Fig. 4(a), 60,000 Q0,000 40. 00 20.000- (b) us 20 4, Inferior span : A 8 8 Q i 4n + 2 4 1 1 *- n 60,000 60. 000 40.00 20.000 OS u^ 20 4i f Exterior spBn ; A 5 i Q* 47T7i li - n n) Fig. 4 rT.L=2200 ,D.l.*IOOQ 20-0 -i ' A/' 20-0 20-0 it • ^ " 20-0 * A U 20-0 20-0 Fig. 5 M AB = M BA = -73,000 + M X 40,000 = -63,000 ft. lb., and Max. pos. mom. = +110,000 - 63,000 = +47,000 ft.lb. To appraise the effect of varying Q-values, the end and center moments given below are determined by both approximate and exact analysis* for Q equal to V20, Y% an d /4> using the load arrangement in Fig. 4(a). Q H ^ l /« n H m 4M End Moment . ... (Exact) -63,000 -60,000 -68,000 -65,300 -71,000 -69,700 Center Moment . . . (Approx,) +47,000 +50,000 +42,000 +44,700 •■ h 39,000 h40,300 The moments are plotted in Fig. 4(b). It is seen that increasing Q from V20 to Y /± (equivalent to decreasing n from 4 J/2 to J^), decreases the end moments and increases the center moment by about 20 per cent. The greatest discrepancy between any corresponding exact and approximate valuo is about 6 per cent or only one-third of the disparity in moment when Q varies from V^ to J^. *"Exact Analysis" in thie text indicates that the frame in its entirety was analysed bv the moment distribution procedure proposed by Professor Hardy Cross (sec Reference 10). The intermediate calculations were recorded to the nearest 100 ft.ll>.. and slide rule was used throughout. Due to cumulate rrors, ti discrepancies in the "exact" values may possibly be as large as *500 ft.lb. 10 The center moment computed by using the conventional coefficient (Hi^P) * s a ^ so plotted and is about 17 per cent greater than the approximate moment determined for Q = M ana< 41 per cent greater for Q = V20. In Figs. 4(a), 5(a) and 6(a), note that two joints and loads on spans adjacent to these two joints only are considered in the determination of beam moments by the proposed procedure. It is merely for comparison of these moments with the exact maximum moment values that the entire frames and all the loads shown in Figs. 4(a), 5(a) and 6(a) are included. ■ Problem 2. The frame in Fig. 5(a) has equal spans, 20 ft. long, and equal story heights; dead load equals 1000 lb. and live load 1200 lb. per lin. ft. Determine the maximum center moment in the exterior span AB and the corresponding end moment M A b. To produce the maximum value of the center moment, the loading is to be arranged as in Fig. 5(a).* Determine the fixed end moments accord- ingly. The beam and column dimensions are not known. Suppose the exterior columns are slender (Q A = }4) and the interior columns of average proportions (Q B = Y%), see Section 3. The calculations are as follows, with moments given in 1000 ft.lb. v>=0 1 A w = 2200, J =20 1 B tt> = 1000, I =20 -73 - l /iiX2200X20 J -VaX2200x20 2 -73 -33 -V12XI 000X20* 1 O-C-73) +73 +37 +2X^X<+73) -KX(+73) -18 - 5 + ^X(-40) -2XHXC-40) +10 -40 (-73)-(-33) -41 M AB Mba -81 The moment calculations in the above table may be written in the follow- ing simplified manner, the computation form in Fig, 2 being omitted: At A: -73,000+2X^X73,000+1 X^X (-40,000) = -41,000 Atfl: -73,000-1X^X73,000-2X^X(-40,000)« -81,000 Average: —61,000 The maximum center moment in span AB equals • +y 8 X 2200 X 20 2 - 61,000 = +49,000 ft.lb. To appraise the effect of varying Q A} values of end and center moments given below are determined by both approximate and exact analysis for Q A equal to M, Vs and V20; Qb remaining equal to J^. Qa H H Vto n A H i« 4}^ . . . (Approx.) . . . (Exact) -41,000 -41,800 -60,000 -60,500 -71,000 -71,800 . . . (Exact) -49,000 h51,000 +44,000 +46,000 +41,000 +42,500 ♦See footnote on page 9. Note that the maximum value of Mab is not produced by the loading in Fig 5(a), but by a loading as shown in Fig. 8(c). 11 The moments are plotted in Fig. 5(b). It is seen by comparing Fig. 5(b) and Fig. 4(b) that the center moments are nearly identical in interior and in exterior spans; their variation when Q varies is relatively small. Com- pared with the conventional moment values of 1 /iqw1 2 for interior and l /\%wl 2 for exterior spans, the moments determined by the analyses will give a more economic design, especially in exterior spans. The end moment at the exterior column, plotted in Fig. 5(b), is more sensitive to variations in the value of Q than any of the other moments plotted in Figs. 4 and 5. Values of Q should therefore be selected with greater care for exterior than for interior joints. The end moments at the exterior column determined by analysis are smaller than those obtained by the use of the conventional moment coefficient of V^. Problem 3. Frame dimensions and loading are the same as in Problems 1 and 2. Determine maximum end moments, M B a, in span AB at first interior and at interior columns. The loading arrangements required to produce the maximum moments are shown in Fig. 6(a).* The analyses that follow are carried out according to the procedure in Problems 1 and 2, assuming for all columns Q = J4, Y% and y 2 o. The results are tabulated below and plotted in Fig. 6(b). The end moment is slightly greater at the first interior column than at the other interior columns; and both moments are greater than the moments computed by using the conventional coefficient of Vi2. 100,000 60,000 T lit Interior column. 60,000 (b) Interior column 40,000- 20,000 5». ^ Approximdte Exact Q- 1 4nfZ In ter'tor column I 1 fa) A 3 1 10 4i 3 li — Q —+ ' I if Interior column n z Fifj. 6 ♦Maximum end moment at B requires (1) full live load on AB, (2) maximum restraint at B. and (3) mini- mum restraint at A; requirement (2) calls for full live load on span adja'i nt to B and (3) calls for no live load on span adjacent to A. Further studies indicate the load arrangement shown above and below floor AB is required to give maximum moment at B, 12 Q M H VfD n H Di 4H First Interior Column, End Moment. . . (Approx.) (Exact) -91,000 -90,100 -82,000 -83,300 -77.000 —78,100 (Exact) -83,000 -86,500 -78,000 -81,700 -75.000 -77,600 Problem 4- The frame and loading will be similar to those in Problems 1, 2 and 3. The value of Q = Y%{n = V/i) will be used at all joints. The span in which moments are to be determined and also alternate spans will be 20 ft. long; for the other (intermediate) spans, lengths of 16 ft. and also 12 ft. will be chosen. In the 20-ft. spans, maximum moments will be determined at the follow- ing points of the beams: Center Interior, Center Exterior, End Interior, End First Interior and End Exterior. The loading arrangements required to produce the maximum moments are similar to those shown in Figs. 4, 5 and 6, except that the approximate end moment in the beam at exterior columns is produced by a loading as in Fig. 8(c) and determined by equation (6). All other approximate moments are determined according to the procedure in Problems 1 and 2, and the results* are summarized in Table 1. Main Span: 20ft Adjacent Span Length Maximum Moments 20 ft. 16 ft 12 ft (Exact) +42,000 +44.700 +43,000 +45,000 +44,000 +45.200 (Exact) H44.000 h46 t 000 +45,000 +46,100 +45.000 +46,200 (Exact) -78,000 -81.700 -73,000 -77,000 -69,000 -72,900 (Exact) -82.000 -83,300 -76,000 -77,900 -70,000 -73,800 (Exact) -63,000 -65.000 -63,000 -65.400 -63,000 -65,800 Tabic 1 The agreement is good between results obtained by the approximate and the exact analyses. The change in moments in the 20-ft. spans is small when the length of the adjacent spans decreases. The maximum change in the center moments and in the end moment at the exterior column is insignificant. The end moments at the interior columns are somewhat more sensitive to change in length of adjacent spans. Moments in the shorter spans will be discussed in Sections 8 and 9. ♦The calculations may readily be duplicated by the reader by using Fig. 2. n Conclusions: For the cases investigated in Pi <lems 1, 2, 3 and 1. the summari. in Flj 4(b), 5(b), 6(b) and in 1 1 show that : (i) The agreement between momenta obtained by the approximat* ml by th« • ; ■ - dun is fiatifif actor (2) Tl ■ moment • pt the end moments at ext< rior lumna, < nj: comparatively little when the relative stiffness, ?.. varies;* (3) .Moments obt ted by using conventional ooeflictanta me % ater (up to about 75 per cent) than moment^ obtained I analvsifl, exoc i that end momenta at interior and fii>t interior columns are small* i (op bout 20 per cent) than th< obtained by analysis. lor neentr 1 Loads winch an symmetrical with reap to tin mid- point of th< joii, the agreement bet 1 en momenta <>bi 1 by the pio- p 1 ami by the i lei procedure WtJ found ^ factory. It is po.vnbl* however, i irranfj. load unaymmetrically with reaped to the midpoints pans that the p ro po s ed procedure max \i ■ unsatisfactory result- but such l<i:i(jinga ao rela ly infi tit. It was r nmended in ■Vet ion 3 that tin minimum valu< lx- el < n one-half. If n illy smaller than this value, using n — J*.- i •■■■• < r ii\ suits for thi that an especially senait <• to \ v narnelv. the moments in columi and in beams at ext< •■>! <<»!- urnm — • -• - !-'■§£ b d 1 ig 9 Other momenta are anVn far !• — I- isll ciinii). and, in general, pi ed no1 l»« takei imalh than on( -half. By the simplified | dun sent I, ma: tnum momenta in continuous buildii fran»es ma> !• letcrmined by w mg a U -u figure.' ud a m< al distribul m oi n tl<nt.- n\uy o|>! tied than b ie convi ill j.iit.' nl e< h 11 lit,*- M«.i at critical po dol ed b) tin ap] imat* method aj rail) f( f <» be than lb « a* f momi n1 ( onaidei ■ t hi i tluixm m I ip < On i he frame i s an i>t« in for < imuni momt nl hut on o ppn \ ,?it< moi I l.i i1 i h\ 2 apaii loading i i 1 |m am all* than !h< fid i monf It hh<»ul< t ! • • rn)>l< tr loading i r i I . I vs ill • st r §trt] c i nt will n e^Hy imium «i. and '.'< I t««d ) (Mttt* b| u*« •aw F»r* « I »vid lviu»' tt> aJJ tipam' A Em m^mtamt at e^rmiBav — - v& 24* • B»d MMBte ia ***** t*mmm li- ft* *tafab work *< «panM ol arrurary b> t^iif 'a 6*f» 14 reached, it is rational to use a smaller moment for design. The approximate moments, therefore, are good design values. 5. Moments in Beams in Roofs and Bottom Floors One column connects with each joint in the roof, while there are two columns for each joint in floors. This difference will be reflected in the expression for Q in Fig. 2, for which the following equation will be substituted : which applies to beams in roofs. Otherwise, the analytical procedure illus- trated for floors and arranged as in Fig. 2 will apply to roofs also. Beams in bottom floors may be analyzed as described and illustrated for regular floors using the value of Q = — (2) * 4n + 2 A part of the regular analytical work may be eliminated when the col- umns are relatively stubby. The calculations involving Q and U (see Fig. 2) may usually be omitted in the following cases: (1) at interior columns whose stiffness ratio n exceeds about 5, and (2) at exterior columns whose stiffness ratio n exceeds about 7J^. The fixed end moments may be used in these cases as final moments, since correction for joint rotation is unjustifiable. 6. Shear in Beams Shear in a beam which is part of a frame is composed of the shear in the beam considered simply supported and a correction due to the end moments produced by the frame action. Shear in a simply supported beam may be determined by statics. For concentrated loads, it is usually satisfactory to use the center-to-center distance between supports as the span in the shear calculations. When the load is uniformly distributed, the shear at the face of the support is one-half of the load on the clear span. An end moment Mab, creating tension in the top of the beam AB in Fig. 7(a) with span length I, produces reactions, R, the direction of which is M A A She3rCur¥e M:0 M JR [ She8rCUrve M 5A »>- dhear curve k 1 M ££/ R= —^ ZL R-- \L M Am *M M / A3 "'BA (a) (b) Vertical load (c) Wind load Fig. 7 15 upward at A and downward at B, the numerical value of the reaction- being M AB /l. The shear is constant and the shear curve is as shown in Fig. 7(a). The shear is also constant when moments are applied at both ends of the beam; but the numerator in the shear formula is the difference between the end moments when their direction is as shown in Fig. 7(b) and the sum of the end moments when their direction is as in Fig. 7(c). Allowance for sign of the shear in Fig. 7(b) may conveniently be made by use of the rule that the reaction on the beam is directed upward at the joint with the larger moment; the direction of the reactions in Fig. 7(a) and (c) may readily be determined. To produce maximum end shear at joint B in span AB, see Fig. 6(a), it is required that (1) span AB is fully loaded, (2) M BA is as large and (3) M AB as small as possible. This indicates the loading shown in Fig. 6(a). For maximum end shear at exterior columns, loads may be arranged as in Fig. 5(a). For illustration, the maximum shear in the end span at the centerline of the first interior column in Problem 3 for n = Y2 (Q = 14) may be com- puted by the approximate procedure which follows, the loading being arranged as on the middle floor in Fig. 6(a), first interior column: from which M B a = -73,300 -lXJiX 73,300 Mas = -73,300 + 2 X l A X 73,300, M AB ~ Mba = % X 73,300 = I 20 Shear at B equals Y 2 X 2,200 X 20 + 2,800 - 22,000 + 2,800 = 24,800 lb. In this case the simple beam shear, 22,000 lb., ha* been increased by 12} per cent at the first interior column and decreased by the same amount at the exterior column. One of the effects of continuity is to transfer shear; and it is often advisable to make allowance for shear transfer in the deter- mination of column loads, particularly when the exterior columns are rela- tively slender. 7. Column Moments and Beam Moments at Wall Columns Maximum moments in columns are produced by placing loads on several beams in at least two bays. The following convenient procedure, which is based on placing loads on few beams only, gives results that are satisfactory. Fig. 8(a) shows a loading arrangement that produces, approximately, maximum moments in interior columns.* Live load is placed on all beams in one baj r and there is no live load on beams in adjacent bays. The unbal- anced moment, U, tends to rotate each joint through the angl< O, and ♦Maximum moment at the cade of a column requires lhat (1) th<- unbalanced moment at t li • ends is maximum, (2) the column deflection curve has reverse curvature, ar '>c rotation of tin id j"inta is maximum. This indicates approximately the loading shown in Fi«. 8 A somewhat target tnomenl >uld l>f produced if live load were placed as shown on two floors on I the loading reversed on the next floor a! rod below. 16 the columns will deflect as indicated in Fig. 8(a). The moments, M, induced in the columns by U are shown in Fig. 8(b). For the conditions in Fig. 8, it can be shown, see Section 28, that the end moments in the interior columns equal M = 3n Fig. 8 Qn + 2 X U, (3) in which n = ratio of K for columns to K for beams (see expression for n in Fig. 9) , U = unbalanced moment due to the vertical loads indicated in Fig. 8(a) . In exterior columns, see Fig. 8(c), end moments may be computed from the following equation: M = 2n X U, . m * * in + 1 in which U equals the fixed end moment at the exterior joint. The procedure is to (1) determine U from the loading given, (2) multiply U by the proper coefficient selected from Fig. 9. • . . (4) The moments thus determined are approximate. To be conserva- tive, it is advisable to choose large values of n and to adopt, say, n = l /2 as a minimum value. The shear, V, in columns with reverse curvature as in Fig. 8 is the sum of the end moments divided by the story height, or in this case V = 2M_ (5) 50 ^30 ^> 45/0 I Ext II \ 2) er/or columns- dn v &n 'nteriorcolumns I It i 12 3 4 Relative column stiffness Fig. 9 8. Since the end shears are opposite and nearly equal in the column above and the column below a joint, the beam between the columns gets little or no axial compression from the loading in Fi Maximum end moment in beams at exterior columns equals the sum of maximum moments in the column above and the column below; it is therefore produced by a loading similar to that which gives maximum col- umn moment and may be taken as M AB = —^- X U, (6) 4n + 1 in which U is the fixed end moment, M^b- 17 Problem 6. A frame has spans and loading as in Problems 1 to 3; values of l A, l l A, tyb are assumed for n. Determine maximum moments in the columns and in the beam at the exterior columns. For the loading arrange- ment in Figs. 8(a) and 8(c), the unbalanced moment, U } is at interior cols. : V12 X (2200 - 1000) X 20 2 = 40,000 ft.lb. at exterior cols.: l / n X 2200 X 20 2 = 73,300 ft.lb. Using equations (3) and (4), the following values are computed: n Interior Columns (Coefficient) (Moment) Exterior Columns (Coefficient) (Moment) \4 1H 0.300 12,000 0.333 24.400 0.410 1 6,400 4H 0.465 18,600 0.429 31 ,500 0.474 34,700 Stresses due to these column moments are illustrated in the following examples: (a) Interior column, n = 4J^; gross section = d 2 = 24 X 24 in.; axial load, P = 500,000 lb. The eccentricity is M 12 X 18,600 n ._ . e = — = ■ = 0.45 m P 500,000 (b) Exterior column, n = 1}4', gross section = d 2 = 16 X 16 in.; axial load, P = 200,000 lb. The eccentricity is e M 12X31,500 tn . = — = - = 1.9 m P 200,000 In a rectangular, homogeneous section with depth d, the extreme fiber stress for load P with eccentricity e is 1 1 + — J times the stress produced by load P applied concentrically. The increase in stress is for (a): '— = 0.11, or 11 per cent, w 24 f or (b); 6 X L9 = 0.71, or 71 per cent. 16 In this example, the effect of moment is negligible in the interior column ; but the moment in the exterior column should not be ignored. The maximum moments in the beam at the exterior column are, using equation (6) for the approximate values: n Moment (Approx.) (Exact) M 1H 0.857 4H 0.948 -48,900 -49,900 62,800 64,700 69,500 70,500 There is good agreement between the approximate and the exact moments 18 Special Procedure of Analysis 8. Assumptions Regarding Restraint at Far Ends of Columns To obtain satisfactory results by the procedure in Fig. 2, the rotation must be relatively small at joints in the floor above and below the floor considered. If the rotations are large, a considerable error may result from assuming that the far ends of the columns are fixed or hinged. The errors that may be large are in the moments that depend upon live load being placed in short spans which are flanked by long spans without live load, this loading distribution being reversed in the floor above and below. In general, such moments are relatively small and not important; but a discussion of the problems involved is desirable and will be given in the numerical problems that follow. Problem 6. The symmetrical frame in Fig. 10 has three bays, the span lengths being 20 ft. for the outer spans and 10 ft. for the center span. The dead load is 1200 and the live load 2400 lb. per lin. ft. All Jv-values will be assumed equal, and the value of n is therefore equal to unity. In order to produce (1) minimum center moment in AB, and (2) maximum center moment in BC, live load is placed on the short beam BC, the general loading being arranged as in Fig. 10(a). The actual condition of continuity is shown in Fig. 10(a), and an assumed condition with the far ends of the columns fixed is illustrated in Fig. 10(b). fa) Actual condition of continuity A 3 C 3 6 -dO.O +20.0 -40.0 -28.2 + 24.8 -42.2 -25.5+25.9 -42.6 -59.9*16,5 -47.2 -28.4+24.1 -43.4 -30.0 +150 -32.2 + 12.8 -33.1 + 113 -25.1 +19.9 -31.6 +13.4 -35.6 +/ 7.7 -48.9-25.5+19.5 A (c) 3 nillllllllllllTTT ^ » 5 n IHIHMMI (b)Far ends of columns assumed fixed "*"» ** »4 //% ■ Actual as In (a ) ■V •* T&6 Proposed procedure fixed enos as in fb) C Fig. 10 w 19 The deflection of the centerlines is indicated by the dotted lines. It is seen that the columns deflect in opposite directions and that the rotations of joints B and C are opposite in the two figures; this indicates that the assumptions made in Fig. 10(b) are unsatisfactory. Moments in the beams AB and BC calculated on basis of various assump- tions are tabulated in Fig. 10(c). Line 1 gives the moments obtained when all joints A, B, C and D are fixed. When these joints are released and the far ends of the columns remain fixed, as indicated in Fig. 10(b), the moments are as in Line 2; Line 3 gives the moments obtained when the columns are assumed hinged at joints above and below ABCD. Compare the moments in 2 and 3 with the exact moments, which are given in 4, and with the moments in 5, which are obtained by the procedure in Fig. 2. The results recorded in lines 2, 3 and 5 are similar; and they are in error, obviously because the joint conditions assumed at the far ends of the col- umns are not correct. For loading and frame conditions similar to those in Fig. 10(a), too great an error may result from assuming the far ends of the columns fixed, hinged or partly restrained. In such cases, proper allowance should be made for the effect of loading on the floor above and below the floor considered. The allowance, which is made in the moments recorded in Line 6, will be discussed in Section 9. 9. Correction for Rotation at Far Ends of Columns In making allowance for rotation of joints above and below floor ABCD in Fig. 10(a), the analytical procedure in Fig. 2 is useful but may be modi- fied as illustrated in Problem 7. Problem 7. The frame, loading and quantities sought are the same as in Problem 6. The calculations, given first without correction for joint rotation above and below ABCD, are recorded in Table 2, which is arranged as Fig. 2. The end moments determined in Table 2 — Mab, Mba and Mbc — are those 1 given in Line 5 in Fig. 10(c). B I y 6 40 -Vu XI 200X20' I, -V.sX1200X20 s +13 3 - 1.7 +2X'«,X(+40) +1X'6X(-10) -1XJcX(+40) -2X'fcX(-10) -28 4 M m Mh -40 -67 + 3 3 -43.4 -30 -33 + 1.7 /nX3600X10' +2 X Hi - 1 0) +1XH(+10) -31.6 .1/ BC Table 2 The values of 40 and 10, underscored in Table 2, are the unbalanced moments at 4, B and C. They are the only values which will be corrected; the procedure of correction is illustrated in the paragraphs which follow. The numerical value of a moment at one end, A, of column A A' induced by a rotation due to an unbalanced moment, V a', at the other end, A', will be taken as n X l\\> X Qa' (see Fig. 11). The sum of moments origi- 20 U.-=(-l?QO) nU A .Q A .'l(tl?0.0)*i" 10. n U.. Q..'lflJ0.0)'i'-l6.3 W--I700 nU.. Q.il(*1200)*i' 1 70. -J-l?O0h' 120.0 nU 6 ,Q r -l(-IIO.O)4-.-m (/*.*(■ naoH-il -no. o »,3bOO 70-0 w-noo Q'( a+ ail joints Fig. 11 nating at the joint above and the joint below A, equalling nU A 'Qa' + nU A >>Q A " f tends to rotate joint A in a direction opposite to the direction of Ua' and U A ». This sum will be included in the determination of the unbalanced moment at A, thereby introducing an allowance for rota- tion at A' and A" . The procedure is illustrated in Fig. 11. The fixed end moment at A' and A" is -120.0, and U r = U A » = - (-120.0) = +120.0. Choosing the value of Q A > = Q A „ = % and inserting n — 1 gives the sum of the moments induced in the column at A : 1( + 120.0)^ + 1 ( + 120.0) Jf = 2(+20.0), the direction of which is opposite to the direction of M AB . The moment tending to rotate A then equals U A = - (-40.0) - 2(+20.0) = . Similarly at column B'BB'\ taking Q B . = Q B „ = % gives U B > = V B » = -120.0 - (-10.0) = -110.0, and nU B 'Q B > + nU B »Q B » = l(-HO.O)}^ + 1(-110)^ = 2(-18.3). The direction of the moments induced in the columns is opposite to that of M BA , which gives U B = -40.0 - (-30.0) - 2(-18.3) = +26.6. In summary, to obtain a corrected value of U at any joint, take the M F to the left minus the M F to the right of the joint minus the moments induced in the columns above and below the joint. The moments induced at the near end of the columns may be taken as nQU written for the far end of the columns, U in this case also being equal to M F to the left minus M F to the right. In the moment calculations in Table 2, replace +40.0 by 0.0, — 10.0 by + 26.6 , and +10.0 by —26.6. The calculations will then be as given in the following table: k -40.0 _i /iiX1200X20= B I _i /12XI 200X20! -40 +2XMXC0) 4- 4.4 -35.6 +1XMX(+26.6) 1/ AB -IXHX(O) -2XJ^X(+26.6) M /*. 1 0.0 - 8.9 -48.9 -30 _i , uX3600X10 2 + 89 - 4.4 +2XK(+26.6) -HX!r,X(-26.6) -25.5 .1/ The center moments equal : in AB: y s X 1200 X 20 2 in BC: H X 3600 X 10 2 M(35,600 + 48,900) = 17,700 ft.lb 25,500 = 19,500 ft.lb. 21 These moments are recorded in Line 6 of Fig. 10(c) and plotted in curve marked "6" in Fig. 10(d), in which they are compared with the exact moments (marked "4") and with the moments marked "2" which are obtained by assuming that the far ends of columns are fixed. It is seen by comparison of the moment curves in Fig. 10(d) that the assumption of fixed far ends is unsatisfactory in this case, and that the proposed pro- cedure gives moments that are close to the exact moments. It should be noted that for a frame as in Fig. 10(a), the general procedure in Fig. 2 — without correction — gives good results for the moments in the floor above and below ABCD. For further illustration, the following cal- culations and comparisons of moments applying to the frame and loading as above are presented for two conditions: (1) all i£-values equal to unity, Q = }4 (moments underscored); and (2) all i£- values equal to unit, Q = %f except for the intermediate beams, for which K = 1/10 and Q — 1/42 (moments given in parentheses). All moments are in ft. lb. Maximum center moment in exterior span. At exterior column: -120,000 + 2 X H X 120,000 + 1 X K(-H0,000) = -98,300 At interior column: -120,000 - 1 XK X 120,000 - 2 X V&{- 110,000) = -103,300 Average: = -100,800 At midpoint of span: Case (1) Case (2) Approximate analysis: +180,000 - 100,800 = + 79,200 (+79,200) Exact analysis : + 82,000 (+84,700) Minimum center moment in interior span. Case (1) Case (2) Approximate analysis: + 15,000 - 10,000 - % X 110,000 = - 13,300 (+2,400) Exact analysis: - 15,600 (+2,500) Maximum end moment in beam at exterior column. Case (1) Case (2) Equation (6): ^^ — - X 120,000 = -96,000 (-96,000) H w 4X1+1 — Exact analysis : - 102,700 ( - 103,800) The center moment of —13,300 ft. lb. under the heading "Minimum center moment in interior span" is produced by the loading shown on the top or the bottom floor in Fig. 10(a). For the same loading, the end moment in the interior span is -10,000 - Ve X 110,000 = -28,300 ft.lb. Since moments created by this loading are negative throughout the entire length of the center span, it will be necessary to detail top bars extending from span AB through BC into CD. The practice of stopping top bars at or near the quarter-points of spans is safe where adjacent spans are nearly equal and live load is relatively mall. When the difference between the lengths of adjacent spans inrreasi 22 and/or the ratio of live to dead load increases, it becomes increasingly important to determine the points to which top bars should be extended, points which may be quite different from the quarter-points. It is recom- mended to arrange the load as for example on the top floor in Fig. 10(a), to compute the corresponding moments ( — 13,300 and —28,300 in the example), and to sketch the moment curve connecting these points. Such curves indicating minimum center moments are of importance in design and detailing of bar reinforcement. In the cases illustrated, it is seen that (1) the accuracy obtained by use of Fig. 2 and equation (6) is satisfactory, and that (2) the change of K for the interior beams from unity to one-tenth has comparatively little effect upon the moments beyond the joints between which the change takes place. 10. Summary The following procedure is recommended for determination of moments produced by live load placed in short spans flanked by one or two long spans without live load. Arrange and carry out the calculations as in Fig. 2 with the exception that the unbalanced moments are modified in accordance with the following steps, (1) to (5), which are illustrated in Fig. 11. (1) Determine Fixed End Moments (recorded to right and left of all joints). (2) Compute Unbalanced Moments, U, at joints A f , A", B', B" (on larger circles) . (3) Determine Q at joints A\ A", B', B" . (4) Compute nQU at A', A", B', B" (recorded above and below A and B). (5) Compute Modified Unbalanced Moment at A and B, to be used in Fig. 2, as M F to the left minus M F to the right, minus the moments induced in the column above and the column below each joint. The calculations as arranged in Fig. 11 are simple and direct; they may usually be recorded without the schematic arrangement. An outline of the analysis for vertical loading in accordance with the general and the special procedures is given below together with references to sections of the text where derivation and description are presented. Vertical Load i ,1 General Procedure I Maximum Beam Moments I End Span at Ext. Cols, M = 4n 4n+1 (Sec. 7) X U All Others 1 Q 4n+2 and Fig, 2 (Sec. 2) I Special Procedure I Maximum Column Moments I I Exterior M = 2n 4n+1 (Sec. 7) X U Interior M = 3n 6n+2 (Sec. 7) X U Beam Moments i i Same as general pro- cedure but with allow- ance for rotation at far ends of cols. (Sec. 9) 23 Dim f/vwon <>/ I arums / //# < [$ \ fr:t! • I m -i die! «i n» t* 1 1 all 1 li< [•ml : « w idl l»> i luti nft m:|UU ill 1 Ur tlih |>l ! \S i Loll £1\ V \ Mil | |i>< MUli, I .' U lid t mtf In lid oth - I In fen | »)t I It .1 1 li «jl I : 1 1 1 < i I |f 111 1.. * «»f lit f fmi in j < ai illH I • ■ I * I ll< ll|l lilt I J I ip! ii I ip 1 1 } I ( i Mil ' i i • Mi> I !i(KMk lit* I •* f 1 I I 1 I n a1 •« pinn :\ - w I it i ix unknown liii< i. I ; i . t' a* i|» « lit fMiii v\ li * dc rd i i p u \ I m I hi ' '1 M li imp! ii y I ) 1< .ii at .id tin nl hi ii u\> < < ! i d 1« r i !1i « i «ll I. nl i ,i I s . f ro) mi liili H r< m , // I n it, u if |( I 1 liftflf . i f < \ ll i i J l *+ ■ j i* i» comparison with moments obtained under certain typical conditions (a), (b) and (c), each case illustrating the corresponding point given above. (a) Assume that the column reaction is triangularly distributed as shown in Fig. 12. The area of the triangle, or the reaction component, is VA 2— J- = %V ' y and the distance from the center of gravity of the triangle to the column centerline is %( - ) = -, The moment of the component with respect to the column centerline is J^Vf r J = x /§Va. This moment, which equals the reduction proposed in point (a), is approximately the amount to be deducted from the centerline moment, M c , when a reaction concen- trated in the column centerline is changed to the triangular reaction shown in Fig. 12. (b) Consider two fixed end beams with the same load, w lb. per lin. ft., the span being I in one, but (I — a) in the other beam. The difference between the end moments in these beams is Vi2^ 2 — \ nw(l — a) 2 = y$wla — l /nwa^. The term ^Inwa 1 is small compared with ' () u-la; by disregarding it and substituting V for l^wl, the difference between the two fixed end moments becomes %Va, which equals the correction recommi oded for the moment at face of column. (c) The slope of the tangent, bf, to the moment curve at b in Fig. 12 equals V and the distance be equals Vl J. The modified momenl curve, which is shown dotted in Fig. 12, lies above the original moment curve between the column faces, the distance between the curves being somewhat less than MF(-) = Y§V a , sav > %( l A>^' n ) = ' aF«. This is the reduction of the center moments recommended due to effect of width of column. In Problem 1, the positive center moment, 47,000 ft. lb. was determined for w = 2200 lb. per lin. ft., and / = 20 ft. assumed to be the distance between the centerlines of columns. Making allowance for a column width of, say, 1 ft. 6 in., the corrected center moment is reduced to 47,000 - r g(2200 X ^°)l-5 = 47,000 - 3,700 = 4: J, 300 ft. lb. The loading in Fig. 6(a) for maximum end moment, 91,000 ft. lb., at the first interior column, B, produces at B an end shear of V = 24.800 lb., computed in Section 6. With a column width of 1 ft. 6 in., the moment at the face of the column at B is reduced to 91,000 - H X 24,800 X 1.5 = 78,600 ft. lb. The question arises whether the beam to be designed is governed by the moment at the face or by the moment at the centerline of the column For illustration, refer to Fig. 13 in which a stress distribution is indicated at the faces of the column. The stress T and a considerable part of the stress C will be transmitted across the column by the reinforcement, but 25 the remaining stresses are distributed over an effective depth which is greater at the centerline than at the face. In European practice, the center- line depth is commonly taken as d -J- -, and the stress distribution is assumed to be as indicated in Fig. 13. The relative increase in effective depth at the centerline is then — -, whereas the relative increase in moment 6a according to Fig. 12 is H Va H X V%wla a Mc-VsVa l / l2 wP - l AX Hwla I - 2a The tensile stresses would be greater at the centerline in cases where the increase in moment exceeds the increase in effective depth; that is, when a ^ a , ^ I — 2a > — or a > I - 2a 6d 6 In general, d is smaller than (I — 2a) /6; and it is then the moment at the column face that governs the design. 12. Effect of Change in Moment of Inertia The research made 1 on the effect of moment of inertia, /, of members in continuous frames, is not sufficiently comprehensive, and some discrep- ancy, therefore, may exist in recommendations for J taken from various sources. In current practice, certain general procedures have been developed which will be briefly discussed. The ratio of K c h X I n = Kb lb X h r in which c refers 10 columns and b to beams, enters into all analyses of building frames. Considerable uncertaintj r exists in determining ratios of Ic (for columns) to lb (for beams). In typical building frames, the entire cross-sectional area of columns i usually effective in taking stresses, and values of / for columns should preferably be based upon the gross concrete section and computed as 6d s /12, in which b denotes width and d depth. It is frequently recommended to add hereto the value of the moment of inertia, I Bt contributed by tin longitudinal bar reinforcement, making the total moment of inertia (I - •) If the column is square and the bars are arranged in acircl<\ tin moment of inertia of the column may be written as: / = % 2 X d* + %a(^ - 1V1 (7) in which d = side dimension of square column, 26 A = area of longitudinal bars, Es Er = diameter of circle through centers of bars, = ratio of moduli of elasticity of steel and concrete. The area, A, in equation (7) is assumed uniformly distributed along the circle with diameter c; and by writing the transformation factor for the bars as — 1, it has been taken into account that the area A has been included as concrete in the first term in equation (7). Cross-sections in beams have both compressive stress and tensile stress. The question arises whether the tensile stresses in the concrete should be ignored. In frame analysis, I is introduced for the purpose of determining slopes and deflections. It is not the particular value of / at any one cross- section, but the /-values over the entire length between joints which govern deflection. Since the uncracked condition generally prevails, the gross area is considered to form the better basis for calculation of I. fed 3 The value of 7 of a T-section is usually taken as - — , in which d is the total depth of the web. The lower limit of b for T-beams is the width of the web and the upper limit is the average value of the panel widths adja- cent to the web; but it is improbable that b can be equal to either one of these limiting values. Professor Cross* has suggested that b be taken as the web width, b r , times a multiplier, C, thus allowing for the effect of the flange. In other words, the moment of inertia of any T-beam may be written as I = CX l-^-l • . (8) 12 in which the quantity in parenthesis equals the moment of inertia of the rectangular web section. For rectangular beams, C therefore equals unity. Values of C for T-beams, computed on basis of the gross (uncracked) sections, are listed in Table 3. The values of 2, 5, 10 and 20 are ratios of "Flange Width" to "Web Width," and the values of 0.2, 0.3, 0.4 and 0.5 are ratios of "Flange Thickness" to "Web Depth." Flange Width Values of C Web Width 1 2 5 10 20 Flange Thickness Web Depth 0.2 1.3 19 2 3 2 7 0.3 1.4 1.9 2.3 2 7 0.4 1.4 1.9 2.4 t 0.5 1.4 2 | t t Average M 1,9 2 3 2 7 Table 3, Values of C *See Reference 6. tThese conditions seldom exist in typical designs. 27 Table 3 covers practically the entire range of flange conditions in typical building frames; and yet, the variation in the value of C is surprisingly small. When the flange width is an uncertain quantity, it is justifiable to consider C a constant, and the value of C = 2 is a good average. Inserting C = 2 in equation (8) gives the general equation for T'-beams: I = y 6 X 6'd», (9) in wmcn j _ momen t. f inertia of JT-beam b' = width of web d = total depth of web If the ratio of flange width, b, to web width, b f , is known, equation (8) is recommended for use with the average values for C given in the last line of Table 3. Equations (8) and (9) may be used directly to obtain values of /, since it is seldom considered justifiable to make any allowance for the reinforcement in beams. When there is more than one beam in a panel for each column, souk designers compute the moment of inertia as the sum of /-values taken according to equations (8) or (9), each term covering one beam web. Not more than one of the parallel beams frames into each column; but — it is contended — the other beams frame into a girder which, in turn, frame into (he column. In this way, all parallel beams in the panel must affect the relative stiffness between the column and the floor construction. It seems more reasonable, however, to include a fraction only of the /-values for beams not framing directly into the columns. If cross-sections are not known when the analysis begins, the designer has to estimate the value of n. It is then important to ascertain how much a discrepancy in n affects the results. Studies of the effect of varying the values of n are given for certain numerical cases in Section 1 Summaries of the results are given in Figs. 4, .") and 6, in which moments are plotted as ordinate*, and the a I issas represent values of n from Yi to V •,. a range which includes most prac- tical cases. The moments that change the most with varying values of n those at the end of the beam at the exterior column. They decreas< from 71,800 for // = \y 2 to 41,800 for n = Yi. Kven in this ca a changi in ti from 4J/<2 to Yl ( i" the ratio of 9:1 > will decn the moment in the rat in of about 5:3 only. For all other moments in Figs. 4, 5 and 6, the effect of varying the value; of // is surprisingly small. It is seen by inspection thai the moment! imputed in Sect ion 4 and given in the accompanying table an not great l.\ affected by relatively large variations in the ratio of f r /Ih ■ n x >> *H M EnH Moment (Interior Column) (First Interior Column) 77,600 (100) 78,100 (100) 86,500(111) 90,100(115) Center Moment (Interior Span), (Exterior Span) 1 40.300(100) 42,500(100) 60,000(124) MOO0 (WO) V •■ moments in th< table, the change with varying value of u ianc from 11 to 24 per nt 13, Effect of Distant Loads In Fig. 2, only a small part of the entire frame is considered in the analysis; this makes the procedure especially convenient to apply. The fact that the analysis may be applied to part of a frame of such limited extent is in a measure due to the smallness of the effect of distant loads. For illustration, let the vertical and horizontal lines in Fig. 14 repre- sent a building frame in which joint A is rotated by an unbalanced exterior moment of 100. No other load will be considered on the frame ; and symmetry is assumed with respect to joint A. The unbalanced moment at .4 will induce moments in the frame, and the end moments induced at the points immediately to the left of each joint on the hori- zontal and vertical lines through A are plotted in Fig. 14 for values of n = Yi^Vi an d 43^. In the horizon- tal line through A, the end moments induced for n = 3 2 are -32.5, +5.7, —1.1 and +0.2 at distances from A of 0, 1, 2 and 3 times the span length. Adjacent to the column through A, the end moments induced for n = 3^ are —32.5, +3.1, —0.3 and 0.0 at distances of 0, 1, 2 and 3 times the story height. The induced end moments become smaller when n increases; they approach zero when n approaches infinity (fixed end beams). At a certain joint in a frame as in Fig. 14, the moments induced by unbalanced moments two or more spans or stories from the joint are small. The rapid rate of decrease of induced moment with increasing distance to point of application of unbalanced moment is a contributing reason why good accuracy may be obtained with a procedure as that in Fig. 2. -32.S Fig. 14 14, Effect of Haunching Haunching of beams will affect the fixed end moments produced by the loading and also the stiffness, which for prismatic beams is expressed as I/l. When analyzing frames for horizontal loads such as wind pressure, the stiffness, S, should be corrected for haunching by multiplying it with a factor which may be selected from Fig. 15 in accordance with the shape of the haunches. For example, let D/d = 2 and a = 0.3 for both ends of a beam AB as shown in the figure inserted in the left side of Fig. 15. The adjusted value of the stiffness S, taken from the left chart in Fig. 15 for D/d = 2 and a = 0.3, equals 2.75(1 d/l), or the stiffness of this haunched beam is 2.75 times the stiffness of a prismatic beam with moment of inertia — Id and span length = /. 29 Fig. 15 For vertical loads, the value of I /I need seldom be corrected for haunch- ing in practical design problems; but the fixed end moments should be adjusted to allow for haunching. For the adjustment, Professor Cross has found it surprisingly satisfactory to use the following simple approximation. For a symmetrically haunched beam with dimensions as shown in Fig. 16(a), the fixed end moment may be taken as (1 + ab) times the fixed end moment in a prismatic beam with the same span and loading. When i he beam is haunched at one end only, the mub lier may be taken as 1 + 2ab) at the haunched end and (1 - ab) at the prismatic end.* Tb- effects of haunching both ends are additive. Problem S. A am AB with fixed ends has unsymmetrical baun< as shown in Fig. 16(b) and carries a uniformly distributed load, w. Deter- mine the fixed end moments. Using the products of ab computed in Fig. 16(b) and the multiple f 1 + 2ab) and (1 — ab), the fixed <nd moments are adjusted for haunch- ing as follows: at A : ( J i2<//- X (1 + 2 X %) X (1 - 1 X 3/40) = 0.116m/2 = 1.39 X x / n wP, at B: C/im'P) x ii - i x X) x ' l + 2 x m = 0.072W2 = o x VW 2 . • to * Pftft 07» | miifif rou* GOtnpftliMOB t ar-t ime&U f<yr tmth «tra L and paral I b«*. exart weao ai>i>T(jxirr-r . .0 The moment is 39 per cent greater at A and 14 per cent smaller at B than the moment, wl 2 /12, for a prismatic beam with the same load and span. The effect of haunching is considerable in this example and should not be ignored. Moments computed by the Column Analogy Method* are 0.115toP at A. O.QSOwP at B. I -3*1 3*b=i * f=£ (a) a*b:fi*i--£ (b) Fig. 16 WIND PRESSURE 15. Introduction A single vertical frame subject to lateral load may be analyzed with a high degree of accuracy by procedures such as the slope-deflection method. In buildings, however, a number of such vertical frames — bents — act together, and the distribution of load to each frame must be determined before the analysis can proceed. This distribution is as important as the subsequent analysis. One of the current approximate methods of analysis is based upon the assumptions that the point of contraflexure is at midpoint of each member and that vertical shears in beams are equal.** These assumptions alone give no indication as to the amount of load carried by each bent. The assumption of contraflexure at midpoints becomes very useful, however, when combined with the assumptions presented in Appendix B. Bents subject to wind pressure have (1) shear deflection which is caused by bending in individual members, and (2) moment deflection, which is due to extension — shortening or lengthening — of column members. Shear deflection is predominant except in tall and narrow, tower-like structures. Moment deflection is ignored in most methods of analysis*** and is also disregarded in the analytical procedure in this text. It should be noted, however, that it is a source of error to ignore column extension, and the error may offset the accuracy of elaborate methods of analysis. ♦See Reference 8. **First put in print in "Steel Construction" by H. J. Burt. Similar assumptions form the basis of R. Flem- ing's "Portal Method" applying to wind pressure analysis, see page 111, Reference 26. """♦Regarding allowance for moment deflection in tall buildings and towers, see Reference 28. 31 A procedure is presented for determination of shears and moments in buildings subject to lateral loading. The derivation — given in Appendix B — is based upon assumptions originally suggested by Professors Wilson and Maney.* The method of distributing wind pressure to the individual bents in this text is similar to one presented by Mr. Albert Smith.** Methods of analyzing building frames for wind pressure will be illustrated and dis- cussed in the numerical examples which follow. 16, Illustrative Problem No. 9 Let Fig. 17 be the framing plan for a floor twenty stories below (he roof, the height of each story being 10 ft. Let the direction of the wind be East- West and its intensity 20 p.s.f. All bays are 20 ft. long. The relative values of /, moment of inertia, and K will be taken as follows: I: Relative Value K: Ratio of / Spandrel beams*** Interior beams . . Wall columns*** . Interior columns . of Moment of to Span Lengt h Inertia or Story Height 20 20/20 = 1 30 30/20 =1.5 40 40/10 = 4 80 80/10 = 8 © f © 4 ® :! © l» * I ,i— N ~* i ii =M-HI ^ ^ 5 II II II ii II II ft — - fr — -a^ == A.__„ © © © © ii ii it ! I ii li ! © © © © © 9.5ff.y ! — J i W ma pressure component Fig. 17 tmnine the distribution of the wind pr< e to tl columns at I he M<mi shon D in Fig. 17. -. . |,:, J] ,r B 21. <ff KclrrenceB >. *** \n B ljuBt it will !»♦• r r in tin.- viilm I / and h for bent J The nine frames from A to J in the East- West direction will assist in resisting the total wind pressure above the floor, which, with 8 bays 20 ft. long and 20 stories 10 ft. high equals W = (8 X 20) X (20 X 10) X 20 = 640,000 lb. Each column in Fig. 17 is part of a frame extending East- West and will carry wind shear. The proportion of the total wind pressure taken by each column is proportional to a "joint coefficient" which is a function of the ^-values of members in the bent. The JC-values must be known or selected before the analysis can proceed. According to the assumption and the derivation in Appendix B, the coefficient at any joint may be determined as follows: Joint Coefficient = (K for col.) X ^ S " m <* *'« for adjacent beams Sum of K's for adjacent members For the floor in Fig. 17 and the iC-values given above, joint coefficients computed by equation (10) are recorded in the second column in Table 4. At joint C3, for example, the i^-value is 8 for the columns, 1.5 for the beam between C2 and C3, and zero between C3 and C4. The coefficient at C3 equals 8 X L5 + ° = 0.69 . 1.5+0+8+8 The next step is to compute the sum of the coefficients for each bent at the floor, a sum that represents the force with which the bent resists a unit translation. The center of gravity of these "bent forces" is then computed by multiplying each force by its distance from bent J and divid- ing the sum of these products by the sum of the forces. The distance from bent J to the center of gravity equals 3458/38.61 = 89.5 ft., the values in parentheses being used for bent J (see Table 4). The distance from bent J is 89.5 ft. to the center of gravity of the bent forces, but 80.0 ft. to the center of gravity of the wind pressure. An eccen- tricity of 9.5 ft. exists wl rich tends to twist the entire building frame with respect to some vertical axis. The twisting or torsional moment may be eliminated by determining, by trial, new X-values for bent J of such mag- nitude that the eccentricity becomes negligible. Bent J is chosen in this case because it is farthest from the center of gravity, thereby giving the change in J great weight; and bent J may be assumed to be a wall in which architectural design may readily be adjusted to suit structural demands. In Table 4, X-values have been trebled for all members in bent J*. By this change, the distance to the center of gravity is reduced from 89.5 ft. to 3458/43.59 = 79.4 ft,, and the eccentricity reduced from 9.5 ft. to 0.6 ft. By adjusting the J-bent and making the eccentricity negligible, the advantage is gained that all joints in the floor get the same translation. This means that the column shears, V A , are proportional to the joint coeffi- > *By stiffening bent J, the beam shear in J (see Table 4) is made twice as large as that in the adjacent bents. The uplift on the windward column, Jl, is therefore twice that on the windward columns in adjacent bents. It is important to ascertain that the uplift in column Jl is not too large compared with the dead load available to counteract it. In case the uplift is too large, the stiffness of members must be adequately decreased in bent J and increased commensurately in the adjacent bents. 3-3 cients in the second column of Table 4, the proportionality factor being equal to 640,000 , , ^ a — ! = 14,i00, 43.59 in which the numerator is the total wind pressure and the denominator the sum of all joint coefficients. At A2 Joint Coefficient 4* 1+1 0*1+4*4 -44 4' 1*1 1+1+4+ d --.BO A3 A4 A5 AG 42 .80 Columns Shear lb- Moment ft. lb. 6,500 32, SO it, 600 same A2 .80 A? .60 At 44 4.08 51 32 33 34 0*1.5 ri 4* „ ,r. f,63 8" 0+15+4+4 Tftt 1 15 +15+8-8 ft.26 82 1. 26 same same Same 59, 000 asjoint asjoint asjoint asjoint Seams Moment ft lb. 65, 000 59.000 A2 A2 A2 At Shear lb. 6,200 5,900 5,900 5,900 6,200 4.08*8*20=653 ^ Joinf Coe fficien t Columns El E2 E3 [-4 Ft F2 9,300 18, 500 same 32 1.26 85 SO 82 1. 26 81 63 6.30 CI C2 CI C4 C5 C6 St .63 32 1.26 6 *l.5+0+8+8 :< * 9 C3 .69 32 1.76 51 .63 same same same 46,500 92,500 93, 000 92,500 asjoint asjoint asjoint asjoint 52 52 82 31 9.280 3.250 9.250 9,250 9.280 F3 F4 630*7*20-882 same same to, too 10. too same same asjoint asjoint 50,500 81 82 I Of. 000 50,500 asjoint asjoint 101. 000 82 61 9.280 9,680 9,680 9,280 Of 02 63 64 HI H2 H3 H4 61 82 81 81 81 81 31 81 82 81 Shear lb. .63 1.26 32 1.26 .63 318 .S3 52 1.26 82 L 26 same same same same Moment ft fb. as joinf Beams Moment ft. fb. Shear lb. Bl 9SJC/nf[ 32 * « as join isjoint 52 81 9.280 9,750 9.280 3.18*4*20: 302 same same same 63 3.18 .63 32 1 26 82 f.26 .63 3.18 .63 1.26 82 f.26 63 3.18 same asjoint asjoint asjoint asjoint 81 82 32 31 9,780 9,250 9.280 3.18*3*20-221 same same same asjoint] 8 1 as join A 82 asjoin 82 same asjoinh 81 9,280 9.750 9,280 3.78*2*20-151 same same same same asjoint asjoint asjoint 81 82 82 asjoinh 8 / 9,780 - 9.750 9,780 3.18x1*70' 16 5.16 516*6*20=620 Jl 01 02 03 04 OS 06 81 .63 same asjoint 82 1.26 same 82 1.76 same **um«*>M 15. 900 A 2 .80 same At 44 same asjoint as joint" 19, 500 asjoint asjoint Bl 82 82 95,000 64.000 A2 At 9,280 9,250 9,380 6,150 6,700 J2 5.47 5.4 1*5* 20-- 54 7 - J3 J4 fA/0.44) 1. 33 19.500 (A2- 0.80) 2.40 35, 300 (A2 ■0.80)2.40 same (Af: 0.44) 1.33 same 91.500 176.500 asjoint 195,00 3 116.500 J2 asjoinn J I 16,580 17,650 18.560 (2.48)7.46 7.46*0*20'00 Sum of alt joint coefficients -43. 59 Moment of joint coefficient! with respect to bent J 3458 Eccentricity if joint coefficients; 600- Jj^| '0.6 ft Factor for joint coefficients = *ffi$< 14, 700 Tabl<- i. Distribution of Pressure to Frames Extending East- West :u Multiplying each joint coefficient by 14,700 gives the column shears recorded in the third column. Column moments are computed as the column shear times h/2, or 5 ft. At each joint, the sum of the column moments equals the sum of the beam moments and will be distributed to the beams in proportion to their iC-values. Beam shear is computed as the sum of the end moments in the beam divided by the bay length, 20 ft. Attention is called to joint D4, where the sum of the column moments, 2 X 79,500 = 159,000 ft.lb., is distributed to the adjacent beams in pro- portion to their K- values as follows : 1.5 1.5 + 1.0 X 159 3 000 = 95,000 ft.lb. to the interior beam, and 1.0 15 Q X 159,000 - 64,000 ft.lb. to the spandrel beam. Table 5 contains a summary of a study made by combining the floor framing in Fig. 17 with columns of varying stiffness. The original X-values, 4 for wall columns and 8 for interior columns, were reduced to 2 and 4 and also to 1 and 2, the floor framing remaining unchanged. For these three designs, the percentage of wind pressure carried by the individual bents was computed, the translation at all joints in a floor being equal. The results are recorded in Table 5. Bent A B C O E F G H J. X-values of Columns T and 2 9.0 14.4 12.4 12.6 8.7 8.7 .7 .7 16 8 2 and 4 9.2 14.4 12.0 12.6 8.7 8.7 8.7 8 7 17.0 4 and 8 9.4 14.4 11.8 12.5 8.7 8.7 8.7 8.7 17.1 Table 5 The percentage of wind pressure carried by each bent is surprisingly uniform despite the variation in the i£-values of the columns. This uniform- ity will greatly reduce the analytical work required for a group of typ- ical floors. 17. Eccentric Wind Pressure If the original if -values for bent J as given in parentheses in Table 4 are maintained, the floor in addition to its translation due to wind pressure will be rotated by a torsional moment equal to the wind pressure multiplied by the eccentricity of 9.5 ft. (see Fig. 17), or M m 640,000 X 9.5 = 6,080,000 ft.lb. 35 The columns resist this torsion and the column shears computed in Table 4 should be corrected accordingly. Mr. Albert Smith, in his paper on "Wind Bracing,"* uses the following method of correction. He lets the torsional moment be resisted by shear at all the joints of all bents in both directions and determines the shear correction, "Tv, in each bent, X, in accordance with the formula 7v = M in which Ix + Ir vxx, (11) X = Bents A to J, r.v = Relative shear in bents X (recorded in Tables 4 and 6), x = Distance from bent X to center of gravity of vx (recorded in Table 6), Ix = v x x 2 Y, v y , y and I Y = similar values for Bents 1, 3, 4, and 6. Equation (11) will, for illustration, be applied to the condition in Problem 9, using the original Jv-values in bent J. Refer to Problem 9 for values of vx and the position of the center of gravity. The calculations leading to vy and y are omitted, but may be duplicated from data in Problem 9. The values of 7. v and 1 Y in equation (11) are computed in Table 6.** Bent r. v X *» = I X Bent Ty X V 2 - ly A B c D E F G H J 4.08X70,5 2 = 20,300 6.30X50.5*= 16.100 ! 5.16X30.5-'= 4,800 5.47X10.5 2 - 600 I 3.78 X 9.5*= 300 3.78X29.5 2 = 3,300 3.78X49.5 2 = 9,300 3.78X69.5 2 = 18,200 2.48X89.52= 19,900 1 3 4 6 6.48X41.32= 11,100 I 2.64X1 -3 s = 00 6.67X18.72= 2,300 2.48X58.72= 8,500 ly = 21,900 I x = 92,800 I x + ly - 114,700 / i - 92,800 Tal »lc6 After having determined Ix 4- ly, the regular procedure would be to compute r.v from equation (11) and then cornet the shear in each column but the following short cut is more convenient. Compute for each bent, -< • Table 7, the value of F = 640,000 , 640,000 X 9.5 »s.61 + 114,700 O) = 16,600 + 53.06c). in which :;8.61 is the sum of all the joint coefficients in Table 4, using •r the bent J. The product.- f F and the joint coefficients in Table l equal the final shears including correction for i entricity. *.-• i Refcrei ;. 1.3. 4 and I oly an in led on the right side of Cable 6. 14 in utumnl th< a ihat t» J' in' lb-South in Fie 17bav«- itiffneae ii ting wind | K nine '•• i<> w a in the North-J- .ill -i. make allowance !• m«l Iculatetl MlTii <Mor a* discuss* <J in - tion 12 U uw* m w\ h th iiian a panel f> ich i olun i ',<> Bent A B C D E F G H J M Ix+I* -(*) 53.0 X( -70.5) 53.0 X( -50.5) 53.0 X( -30.5) 53.0 X( -10.5) 53.0 X(+ 9.5) 53.0 X( +29.5) 53.0X(+49.5) 53.0 X( +69.5) 53.0XC+89.5) -3,700 -2,700 -1,600 - 600 + 500 +1,600 +2,600 +3,700 +4,700 + F = 16,600 M Ix + I (x) 14,700 12,900 13,900 1 5,000 1 6,000 1 7,1 00 1 8,200 19,200 20,300 21 ,300 0.88 0.95 1.02 1.09 1.16 1.24 1.31 1.38 Table 7 The last column in Table 7 gives ratios of F/14,700, which is the ratio of column shears with and without eccentricity, see also Table 4. When the eccentricity is 9.5 ft., the shear varies considerably from bent to bent, the shear in bent H f for example, being 38 per cent greater than it is when no eccentricity exists. It is seen that changing the stiffness of a bent may greatly affect the distribution of wind pressure; and for this reason, values of K should include allowance for effect of haunching as discussed in Section 14. 18, Accuracy of Approximate Analysis In order to study the accuracy of the results obtained by an analysis as applied in Section 16, two frames were analyzed for wind pressure. The first frame is one which Professors Wilson and Maney have analyzed by the slope-deflection method;* the second frame has been analyzed by Mr. John E. Goldberg** by a method of converging approximations. The results of the analyses are given in Fig. 18 for the third floor only; they are typical for most of the floors in the bent. The moment values in parentheses in Fig. 18 are taken from the sources mentioned. Moments obtained by the procedure used in this text are under- scored, and typical computations are as follows. The joint coefficients in Fig. 18(a) are 21 4 at exterior columns: 35.4 X tt— — : — ^— — — r^— : = 8.20 21.4 + 35.4 + 35.6 at interior columns: 35.5 X ' ' 21.4 + 29.2 + 35.5 -f- Sb.b Sum of coefficients for four columns: = 14.75 45.90 The column moments, in in. lb., are obtained by multiplying the joint coefficient by 6,690 X (7 X 12) 45.90 7,140 X (8 X 12) 45.90 = 12,240 for columns above 3rd floor = 14,930 for columns below 3rd floor, in which the values of 6,690 and 7,140 are the wind shears given in Fig. IS *See Reference 21. **See Reference 34. 37 The moments in beams are obtained by distribution of the sum of the column moments; thus, the end moment in the center span is determined as (181,000 + 220,000) X 29.2 29.2 + 21.4 = 231,000 in.lb. 6.690 5^> (90.1) 3 r -4 Floor* 222 (203.0) 2/4 12? * v42 0) 032.0) no I 23/ (184.0) I (226.0) 220 X (d) (226.5) 6.69 * ** 18-0 t **r 22 : 115 (109) 3 c? 'floor I 256 (226) % 206 (211) 195 I 266 (206) I (252) 253 '9.2 (258) (b) Fig. 18 Figures circled in are values of j Other figures are moments at joints due to the wind pressure indicated. figures underscored are obtained by approximate procedure. Fig ure s in paren thes is a re computed by: fa) Wi I son-Maney(Reference2l) (b)Ooldberg (Referenced) 18-0 (172) 262 I 1/5 (250) I (122) 207 (210) The frame in Fig. 18(b) is the same as that in Fig. 18(a), except that the stiffness of Columns C and D and the intermediate beam has been reduced as shown. The procedure in determining the moments underscored in (b) is the same as that used for moments in (a).* According to studies made by Mr. Albert Smith, moment values such as those underscored in Fig. 18 may be corrected by the following procedure : (1) compute a correction equal to one-fourth of the difference between the larger and the smaller moment at the ends of a member (beam or column), (2) deduct the correction from the larger moment, and, (3) add the cor- rection to the smaller moment. Consider, for illustration, the end span in Fig. 18 (a) and compute: (1) the correction % (222 - 170) - 13, (2) the corrected moment to the left 222 - 13 = 209, (3) the corrected moment to the right 170 + 13 = 183. By such corrections, some allowance may be made for the fact that the point of contraflexure is not at the midpoint as assumed in the derivation. *The moments in parentheses in Fig. 18(b) do not satisfy the requirement that their sum equals tero at each joint. At column A. for instance, the sum of column moments, 109 + 133 = 242, should equal the beam moment, but the beam moment is given as 226. These moment values are approximate. 38 19. Moment Deflection and the Ideal Layout The beam shears in Table 4 are practically equal in all spans of a regular bent. In such bents, the sum of shears is small at the interior columns, and these columns will receive little or no direct load from wind pressure. The direct load in the exterior columns equals the shear in the end span and is about 9,300 lb. in bents, B, C and E to H, the direct load being tension in the windward and compression in the leeward column. The result is a differential column extension and a warping of the floor, which causes a "secondary" distribution of shears and moments. It is possible to arrange a structural layout so as to minimize the effect of differential column extension and thus approach an "ideal layout" which may be accurately analyzed. In the ideal layout there is (1) no wind pres- sure eccentricity and (2) no warping of the floors. The building frame should be laid out with symmetry in both directions; or, at least, the wind pressure component should go through the center of gravity of the joint coefficients, a case which was illustrated in Problem 9. To avoid warping the floors, the interior bays should be designed to carry much larger shears than the end bays. This may be accomplished by making the coefficients at exterior joints sufficiently small compared with the coefficients at interior joints. To illustrate the procedure in minimizing the warping of floors, A'-values for bent B in Fig. 17 and Table 4 will be adjusted as follows and the wind pressure distributed accordingly, the calculations being recorded in Table 8. Span BI-B2, B5-B6 B2-B3, B4-B5 B3-B4 X-value 1.0 ' 1.5 1.7 Bent B is assumed to receive the same total wind pressure, 92,600 lb., as in Table 4; and shears, moments and direct column loads are computed in Table 8 by the same procedure as used in Table 4. Joint Joint Coefficient | Columns Beams Direct Column Load Shear Moment Moment Shear Adjusted Original 1 m 0+1 /\ v — n AAA 7,200 36.000 72,000 7,100 +7,100 +9.300 0+1 +4+4 B2 8X 1+^+8+8 = 1 " 080 17,500 87,500 70,000 105,000 +3,220 1 0,320 B3 HV 1 - 5+1 - 7 1333 21.600 108,000 101,300 114,700 +1,150 U X 1 .5+1 .7+8+8 _1 "" J 11,470 £4 1.333 21,600 108,000 114.700 101,300 -1,150 10,320 ■ B5 1.080 1 7,500 87,500 105.000 70,000 -3,220 7,100 56 0.444 7,200 36,000 72,000 -7,100 -9,300 Tablt- 8 M) The direct column loads determined for the original and also for the adjusted K- values (see Table 8) are plotted in Fig. 19. The curve marked "ideal" in Fig, 19 indicates the magnitude of direct column loads that will cause no warping of the floor and, therefore, no secondary distribution of moments and shears. It is seen from the curves in Fig. 19 that a much better condition for accurate wind pressure analysis is created by selecting pro- portions of beams in accordance with the K- values suggested. From inspection of the joint coefficients in Table 8, it is seen that the adop- tion of new column sections may be similarly effective. It is especially advisable to make the wall columns, Bl and B6, rectangular with the smaller dimension in the East-West direction. The effect of warping of floors becomes increasingly serious the shorter the outer span is compared with interior spans. In such cases, warping may be minimized and the shear reduced in end beams by judicious Selection Of Shallow OUter ^^ Mi^tes are direct column load due ro»ind. beams and narrow exterior columns. Fig. 19 H.(i>00 1 1 00 9.300 20. Illustrative Problem No. 10; Vertical Load and Wind Pressure Combined The problem is in two parts: (A) Analysis of bent G in Fig. 17 for vertical loads, and (B) determination of maximum moments and shears due to vertical loads and wind pressure combined. (A) Analysis for Vertical Loads. Bent G in Fig. 17 has three spans, each of which has a length of 20 ft. from center to center of the columns; the width of the bay carried by bent G is 20 ft. The successive steps in the analysis will be the determination of (a) Loads. (b) Values of K or n. (c) Values of joint coefficients and moment coefficients. (d) Fixed end moments. (e) Moments at ccnterlines of columns. (f) Moment reduction for effect of width of columns. (g) Shear. (h) Maximum values required for design. (a) The loads in lb. per lin. ft, of the beams will be taken as Dead Load (D.L.) Live Load (L.L.) Total Load (T.L.) 75 X 20 + 300 100 X 20 = 1800 2000 3800 hi (b) The A'- values* will be the same as those used in the wind analysis of Bent G (see Problem 9, Section 16) : Wall Columns: K Interior Columns: K Interior Beams: A' = 4 = 8 = 1.5 (c) Values of joint coefficients and moment coefficients based upon the A-values in (b) are recorded below, see summary in Section 10 for equations and references: Joint Mark See Fig. 20 ~TT =n Beam Moments 1 Column Moments Value of Q At Ext CoL Exterior Interior 1 4n 4n+1 2n 4n+1 3 71 6n+2 4n+2 G\ G1 4 T.5 8 1.5 0.079 0.043 0.92 • ■ m m 0.4$ m a ■ a 0.47 (d) Fixed end moments, in ft. lb., using in Fig. 1, are** D.L.: -Via X 1800 X 202 = L.L.: -Via X 2000 X 20 2 = T.L.: -V I2 X 3800 X 20 2 = the coefficient in case 5 -60,000 -66,700 -126,700 (e) Moments at centerlines of columns, computed according to the pro- cedures in Sections 2 and 7, are numbered below. Corresponding numbers are written on Fig. 20 to indicate the point where the moment is taken and the loading arrangement which produces it. 03 04 01 02 03 64 01 62 63 64 * 8 9 10 & Fig. 20 1. 2. 3. 4 Maximum end moment in G\ —G2 at Gl : M x = 0.92 X (-126,700) = -116,500 Maximum column moment at Gl : Af 2 = 0.46 X 126,700 = 58,300 Maximum column moment at G2: Af 3 = 0.47 X 66,700 = 32,00 Maximum end moment in G1—G2 at (72: M 4 = -126,700 - 1 X 0.079 X 126,700 136,800 ♦When cross-sections are not known, select values for in accordance with the classification suggestc I in Section 3. If the cross-sections are known, values of I — and iC— may be determined by using equations (7), (8) or (9) in Section 12. **For numerical example involving unequal spans, see Problem 7. 41 5. Ma ouni ctnt< r in«>m<i)t Gl — < * • M:-l 0+2X0.079X1 0+1X0.043X (-66,700) « — 109,000 at < -1 jki- 1XQ( >X\ 7Q0-2X <»"1 -6< ,:<hi «_- 1 ;u,100 age: - 120,300 3/ 5 - + X 8800 X . - 120,300 - ■ 6. Mai m center d n< in G2 * d t -12 00 + (2X0.079 IXOJ X 06,700 - - J J 1,400 M 6 - + x 3800 X KP - 121,100 - +68,600 l jim I a: G] to t>< used for maximum shear at U, - i 700 + 2 x 0.078 X 128 - 106,600 (f; M i effe of width oi oolunu - follom, ' iiiiiu *. hmoii l hum' 1 ' < i iii inil< on from Gl i < < l HI (MM) s linn 1) -1/, 1 I f Iaj X 3SiMiit x : ■ — M »o lf< - • ) - ';< X 3£ Kl X 2 - - I" Ml aft -4 • - ' X 88,000 X : - H 89, mi U„ - + <;v ,i! - i , x 000 x - 4 hi < at tin- far In .lu it, tl unlll. f Wl t) if ' • ' : • ' ' - < "luptlted I f<ll|c>W>.*** 1 ll«' 1 1 1 1 f 1 1 1 •< T- iK'loW U f«TM|l| »«- I- ii I 10 a it li load < <i i . i ma iun la. 4 d Gl 1 « S00 X (KM) |. M I iilM -ii at 1-380,.' ! " l: ,,.„,,, I < ...» M «hm 3 at t J/y - 121 I / U H , » -1 X 66,700 k \i r roo - - - i I • » it i/. mm »• - I (h) Maximum* values required for design are as follows: Moment in beams: M lt M i} M 5 , M & . Shear in beams: F 4 , V$, Fg. Direct compression in beams is disregarded. Moment in columns: M 2 , A/ 3 . Shear in columns is disregarded. Direct compression in columns taken from the customary computations of column loads needs adjustment because the frame action transfers some column load from Gl to (?2. An increase in column load of 5 per cent (see F4 = 33,300 + 1,700) at (72 is conservative, but no decrease in column load at Gl is warranted in this example. (B) Maximum moments and shears due to vertical load and wind pressure combined. The frame is the same as that analyzed in (A) for vertical load and in Problem 9, Section 16, for wind pressure. Using the centerline moments determined in Table 4, the beam moment in ft.lb. at the face of all the columns in Bent G is m w = * i ::■::: ) x --—— = * si, 000 The maximum moments, Mi, M4 and M 5 , in the end span computed in (A) are plotted in Figs. 21(a) and (b) and connected by a parabola, which will be considered the curve of maximum negative and maximum positive moments due to vertical loading.** Beam moments due to wind pressure are indicated by the lines connecting ±8 1,000 with =f 81,000 in Fig. 21(a) and (b). The combined moments are measured between the inclined line and the parabola. The maximum positive and maximum negative moments taken from Figs. 21(a) and (b) are plotted in Fig. 21(c). It is seen that wind pressure added to vertical load increases the positive moments and also the negative moments at both ends of beam GI-G2; the length in which there may be tension in the bottom or tension in the top of the beam is also increased. The maximum shears, F 4 and V&, are plotted at Gl and (72 in Fig. 22(a) and (b) and connected with a straight dotted line which will be considered the curve of maximum shear due to vertical loading.** The shear in the beams at the face of the columns due to wind pressure is constant and equals F. = 81fi0 ° + 81 ' 00 ° = 9,300 17.5 The solid, inclined lines in Figs. 22(a) and (b), drawn at a constant distance of 9,300 from the dotted lines, represent the combined shear. The maximum shears taken from Figs. 22(a) and (b) are plotted in Fig. 22(c). It is seen that the shear may be increased over the entire length of the clear span. ♦For discussion of minimum values of moments and their significance in detailing bar reinforcement in frames with unequal spans, see page 22 in Section 9. **In the intervals between the points plotted, moments and shears may be numerically larger than those indicated. The discrepancies are frequently small and will be disregarded in this ease. 43 01 (3) Direction of Wind *59.IOOtMr) 02 -61.000 *32, 100 (a) Direction of Wind mooo -64.800 (M t ) (b) Direction of Wind 59. 100 (M s ) fy D irection of Wind 02 CO Max. combined moments *59.IJfO(M<) 01 M +4l,40d^Max. combined shear w t -35, 000 (y 4 ) 02 Fig. 21 Fij£. 22 Tin- principles embodied in a comprehensive analysis for wind pi sm have been presented, discussed and illustrated in the foregoing rliuri; For the Bake of simplicity, the examples have been prep d for compara- tively simple structural layout. For a frame with highly intricate wind ■lion, Mr. Albert Smith has published a detailed solution for one floor, Hcferen* 56 in the bibliography. In the solution on pages 919 and 920 — it should be observed — the calculations for joists and Mush beams might have been lumped in an approximate total without ther v intro- ducing appreciable error in the final moments in the important I in More than one-half of the calculations given might then have been omitted ti APPENDIX A: DERIVATION OF FORMULAS VERTICAL LOAD 21. Angle Changes Determined by Moment Area Principle A prismatic member connecting two joints, A and B, will deflect when loaded as illustrated in Fig. 23. The angles between the line connecting A and B and the tangents to the elastic curve at A and B will be denoted as Q A at A and 6# at B. The relationship between these angles and the loading may be established by the moment area principle.* Let ACDB be the moment curve due to the loads, P, placed on the simply supported beam AB. Apply the cross-hatched area — the moment :irea — as vertical load on a beam similar to AB and determine the end shears, V& and V&, produced by this Loading. According fco the moment area principle,** the angle changes in Fig. 2'A are <>i = ,!, X I .t.and B B = ,| T X V B in which EI E = modulus of elasficitv and EI (12) / = moment of inertia of the cross section of .1 B, A Pc Pr 3 Fig. 2.'{ Equation (12) expresses that "the angle changes at .1 and B in Fig. 23 equal (1/EI) times the end shears due to the moment area applied as load on beam AB." This is a form of the moment area principle, from which the procedure followed in frame analysis may be developed. It is customary to define signs for end shear due to vertical loading as positive at the left and negative at the right support, and the same con- vention applies to angl< ■ hanges. Further discussion of sign conventions is given in Section 27. Problem 11. (a) A simply supported beam, ^.5„has a load, w } uniformly distributed over the entire span length, I. Determine the angle ehang< at A and B. *The premutation of the moment area principle ilates bark to Otto Mohr who, in I vbUahi " method of computing deflections as if they were bending momenta The lat- (' E Greene at ti University of Michigan, about 1873, discovpred a related principle of moment! m. H. Mull>T-Breslau. 1885, gave Mohr'a principle a general formulation. ♦♦Derivation of the principle is given in many textbooks =>uch aa Referen I (p 10) and Reference 1 l (p. 210). 15 ID I The moment curve is a parabola with a maximum ordinate of — . Load 8 beam A B with the area under the parabola which equals % X determine the end shears : T and V A = -Y B = MX y 3 x wl s wl z b 24 The angle changes, according to the moment area principle, equal o A = -e B = 24JET * • * • (13) (b) A concentrated load, P, is placed on beam AB at a distance of al from A and bl from B; show that the angle changes due to the load, P, are A = ~6EI ( ' an = PabP 6EI (1 + <0 (14) 22. Angle Changes Due to End Moments The loads on beam AB in Fig. 24 are two end moments, M AB and M BAr which deflect the beam as indicated. Let the angles between the line AB the tangents to the elastic curve at A and B be denoted as Q A at A and Q B at B. The relationship between these angles and the moments that produce them may be established by the moment area principle. Fig. The moment curve is the straight line connecting points with ordinate* M AB at A and M BA at B. Load beam AB with the cross-hatched moment area ACDB in Fig. 24, divide the trapezoid into two triangles as shown, and determine the end shears: v, = .' i; = B (HMab xi)xy 3 + (y 2 M BA x o x H = HMab X / + HMba x i -('AMab X I) x M ^ ab X I Wba X J) x h BA X I . (15) According to the moment area principli xpressed in equation (12), the angle changes 8' in Fig. 24. derived from equations (15) by selling 1 0' = — X V\ are hi Q'a = J^ X (23/^ + V flA ) n, = X (2Jl/« x + 3f iB ) ( 1 ' 16 Equations (16) give angle changes in terms of end moments. This relation- ship is important in frame analysis. 23. End Moments Required for Fixity To determine the fixed end moments, Ml B and M£ At due to a load, P, consider first the beam AB simply supported as in Fig. 23, the tangents being rotated through angles expressed by equations (12). Then apply, as in Fig. 24, two end moments of such magnitude that the tangents at A and B are brought back to their original position. The latter step is equiv- alent to fixing the two ends of beam AB, and the end moments, M% B and M BA , equal the moments at the supports of a beam with fixed ends. The procedure is expressed algebraically by the equations B A + 9^ = and 05 + Qb = 0, the 9- values being taken from equations (12) and (16). The equations become V A - -J(2M & + Mg A ) V B = +|(2M& + M F AB ) 6 F Solving for M* gives M F AB = -|(27j + V B ) 2 > (17) MSa = +~(2V B + V A ), s in which V A and V B are the end shears due to the moment area for load P applied as vertical load on the simply supported beam AB. Since V A is positive and V B is negative, the fixed end moments are negative quantities. Further discussion of sign conventions is given in Section 27. Equations (17) are used for determination of fixed end moments, such as those presented in Fig. 1. Problem 12. (a) A beam AB has a load, w, uniformly distributed over the entire span length, I. Determine the moments, M AB and M BA required to fix the ends of AB. Inserting in equations (17) the values V A = -V B = U ^ A B 24 as determined in Problem 11 gives M F AB = M F BA = -^ (18) (b) A concentrated load, P, is placed on a beam AB at a distance of al from A and bl from B; show that Mi B = -ab 2 (Pl), and Mg A = -a?b(Pl) (19) Equations (18) and (19) are used frequently in frame analysis; they should be memorized for convenience. 47 24. Moments as a Function of End Rotation and Fixed End Moments If joints A and B in Fig. 25 are locked artificially before the loads P are applied, fixed end moments — M% B and M$ A — will be created which tend to rotate the joints. When the artificial restraint is removed, the joints will rotate through angles, Q A and B in Fig. 26, until equilibrium is P. M 8A ; -)( Joints A and 3 locked 3 Joints A and 3 released** Fig. 25 Fig. 26 attained. New end moments, M' AB and M BA , are thereby induced in AB in addition to the moments M F that already existed. The final moments are then M A b = Ml B + Ma B} and M BA = M| A + M BA - : - (20) The values of the moments induced by the angle changes may be deter- mined from equations (16) by introducing M' as the moments induced by the angle changes, 0: 2M' A b + M BA = GE X - X e A M' AB + 2M BA = -6E xjxOi,. Solving for M AB and M BA gives Mar = itm* = 2E X ~ X (26 A + Q B ) ) / . . „ , (21) which inserted in (20), - being denoted as K, gives the final moments Mab = M f ab + 2 X (2EKB A ) + 1 X (2EKQ B )\ M BA = Mg A - 2 X (2EKQ B ) - 1 X (2EKQ A )f . . (22) Equations (22) are the Sloj -Deflection Equations; they express moments in frames in terms of M' and joint rotation. Joint translation i disregarded in equations (22) in accordance with common practice Equations (22) express thai an end moment, M AB} in a beam .1 H whh-h is part of a frame, may be determined as the algebraic sum of I jiree terms: (1) fixed end moment I A: M ABi (2) moment induced by joint rotational t h<- near end, A : 2(27:' A' o., ), and (3) moment induced by joint rotation at the far end, B: \(2EKO B ). Fixed end moments may be determined when loads and spans ao trivr q The vain of EK& used in (22) will b< giv< □ furl her stud- 18 25. Joint Rotation a Function of Stiffness Fig. 27 shows four members which are part of a frame and intersect at joint A. Let joint A be rotated through an angle, 9^, by a moment, U Af applied at A. At the far ends of the members, four different conditions are assumed for illustration: joint B is fixed (Q B = 0); at joint C the rota- tion is —e A} at D it is +9a, and at F it is fQ A , the value of / depending upon the conditions at F. The y -values will be denoted as K ABi K ACt K AD and K AF , which represent the stiffness of the various members. The joint rotations in Fig. 27 have induced end moments which may be expressed by use of equations (21). Since joint A is in equilibrium, the sum of the end moments at A must equal U A . This requirement is expressed in the summation that follows:* M AB = +2EK AB X (2G A + (0)) M AC = +2EK AC X (29^ + (-9 A )) M AD = -2EK AD X (29 x + (+0J) M AF = -2EK AF X (29 A + (fO A )) = +[2EK AB (2e A )} = +[2EK AC (B A )) = -{2EK AD (3 G A )} = -[2EK AP (2 + f)9 A ] 2M AX = U A = 2Ee A X [2K AB + K AC + 3K AD + K AF {2 -f /)]. M A b %**>% M*£25JWA F e B *o (f* 0) M AC =K AC (2E0 A ) 2BQ A - ' A UK AX *(2*f x ] in which x is 3. C DorF M A F*K„f*f)(2£9 A ) (d) \ M AD Moments induced by U A f2FQ A ) /[& =+& A Fig. 27 The four moments induced by U A are shown diagrammatically in Fig. 27(a). Since all four moments tend to rotate joint A in the same direction, it is the sum of the quantities within the four brackets above the summation line that equals U A . *For discussion of signs used in the derivation of equation (23), see Section 27. 49 The equation XM A x = U A gives U 2E ° A m 2K AB + K AC + ZKad + K AF (2+/) ' ' ' (23) Equation (23) expresses that 2EQ A equals U A (unbalanced moment due to the fixed end moments at A, see definition in Section 27), divided by the sum of the products K X (2 + /) written for each adjacent member. By comparing equation (23) and Fig. 27, it is seen that 2 + / = 2, when angle change at far end equals : / = 2 + / = 1, when angle change at far end equals — Q A : / = —1 2 +/ = 3, when angle change at far end equals +Q A : f — +1 Equation (23), written in its general form, becomes ^'- zgJW (24) 26. Derivation of Formulas for Joint Coefficient, Q A quantity denoted as Q is introduced and defined in Fig. 2. It will be seen by comparing Fig. 2 with equations (22) that the value of Q A is An expression for the right side of the equation is obtained by multiplying with —y2- in equation (24), which gives Qa = 2EK AB Q A K A b /ggx U A X[K AX (2+f x )} Therefore, Q A is a function of K A x'- the I/l- values of the members intersecting at joint A, and fx : the degree or type of restraint at the far end of the members. An approximation is now introduced by inserting in equation (25) : 2 + fx ~ 2 for all columns, 2 -\- fx = 2 for beam adjacent to exterior joints. 2 + fx — 1 for beams adjacent to interior joints. If, in addition, the stiffness of beams adjacent to any one joint is set equal to K and the stiffness of columns at the same joint equal to nK, equation (25) may be written in the simple form : Q = — — .** (26) * 2(2tl&) + 2K 4n + 2 Either equation (25) or (26) may be used to determine Q. The procedure in Steps (1) to (6) in Section 2 would give mathematically exact results *A quite accurate and very quick analysis may be made by designers who are accustomed to judging the relative magnitude of joint rotations. The procedure is simply to estimate the value of /for the far ends, determine joint rotations from (241, and compute the final moments from equations (22). Professor Maney has developed an approximate method involving the following steps: (1) at alternate jointB set/i = and calculate 9 from equation (24) ; (2) using these 0-values, calculate at the intermediate joints; (3) re-calculate G at joints in firstgroup; (4) repeat steps (2) and (3) if necessary: (5) calculate moments by equations (22) . This method was presented by John E. Goldberg, see Reference 9. A similar procedure has been presented in a manuscript by Raymon C. Buell. **For roofs, the equation becomes Q - = — iTnt based on assumptions similar to those made for floors. 50 if / in (25) were known; it becomes an approximate method* because / is estimated or approximated as in (26) . The value of /appears always together with K in the product of K(2 + /) . The i£-values are not known beforehand, but must be estimated before a frame can be analyzed; it is then reasonable to estimate / also and to use the simplified equation (26). When reviewing a frame with given pro- portions, the "physical characteristics of the structure make it impossible to determine the actual bending moment to a greater degree of accuracy than =*=5 per cent."** Therefore, even when proportions are given, a rela- tively poor estimate of/ is of little consequence. 27 * Conventions and Use of Signs Sign conventions used in current practice and their application to the derivations in the foregoing analysis will be discussed. A beam, AB, through which an arbitrary vertical section S-S is drawn, dividing the beam into two parts { M" and "f?," is shown diagrammatically in Fig. 28(a). Remove "B" and add to "A" the internal stresses, C, T and V, exerted by "B." The stresses C and T form a couple, the internal moment; since "A" is in equilibrium, the internal moment equals the couple formed by the loads and reactions on "A," the external moment. Trans- posing the procedure by removing "A" and adding in S-S the internal stresses exerted upon "B" gives an internal moment which is numerically equal to the internal moment on "A" but opposite in direction. DOS Points of contra flexure Fig. 2ft Fig. 29 The internal moments at any section may be indicated by two curved arrows as in Fig. 29(a). If the section is taken immediately outside joint A in Fig. 29, the arrow which is closer to A indicates the moment with which beam AB tends to rotate joint A, and the other arrow indicates the moment exerted by the joint upon the beam. ♦Various analytical procedures have been proposed in which a relative stiffness such as K X (2 + /) is introduced. See, for example, Professor L. E. Grinter's discussion of "A Direct Method of Moment Distri- bution." Proc. A.S.C.E., March, 1935. page 426. ♦♦Professor L. E. Grinter in Proc. A.S.C.E., March, 1935. page 428. 51 The internal force V, the shear, in section S-S in Fig. 28(a) is defined as the algebraic sum of the reaction, R' } and the load, P'\ on the part of beam AB to the left of S-S. The common convention is to take the shear at the end of the beam as positive at the left support as in Fig. 28(b); the end shear is then negative at the right support. Apply this definition to the type of end shear treated in equations (12) and (16) ; for a deflection as in Fig. 23, becomes positive at A and negative at B, and these signs are reversed for a deflection as in Fig. 24. From the definitions of shear, it follows that fixed end moments are negative (see discussion of equation (17) in Section 23). The sign conventions for shear and moment are inter- locked by the moment area principle expressed in equation (12). Since fixed end moments for vertical loading are negative and occur where the elastic curve is humped, "a moment is positive when it sags the beam;" this will be referred to as si^n convention number (1). For columns, the ime sign convention may be applied if the sheet is turned to read from the right. Positive moments will then create tension in the bottom of beams and in the right hand side of columns. In dealing with joint rotation, it is convenient to adopt another conven- tion, number (2): a moment exerted upon a joint by a beam is positive when it tends to rotate the joint in the clockwise direction. * Convention (2) would make all the moments at joint A in Fig. 27(a) negative. This sign convention is preferred in analysis, the determination of rotation and moments. The ultimate purpose of analysis is design, or determination of stress ; and for this purpose it is customary to use sign convention (1). Fig. 20 shows deflected members in part of a frame, and internal moments arc indicated by double arrows. The moment equals zero at the two points in AB where the curvature changes from a sag to a hump, the points of contraflexure. The moments, positive between and negative outside tin points, according to convention (1), may be plotted indicated in Fig. 29(b). If convention (2) were adopted, the momcnl in Fig. 29(a) would become positive at A but negative at B; this would be confusing in plotting moment curves. Convention (1) is adopted in this text, although l tain derivations may appear less direct; but th< application of formulas and procedure has doubtlessly been simplified. The numerical value of r A is defined as the algebraic differ* ace between the fixed end moments at A. According to equation 2 1), V has the same BlgQ as &. This requirement will be fulfilled in Fig. 2 when U at an\ point is taken as the algebraic different. U M ! to the left minus M r to the righl of the joint," the fixed end moments being w< d with their prop< n- If at joint A t for ex mple, M AJi > s numerically larj. than M\,, then U A = M', - M is p itiv and o also become- positive; if Mis ] " maller than \1 \, . both Ua and B A are negative. K'.fereii'i ], page 14 52 28. Formulas for Moments in Columns Let K c and Kb be the J/Z-ratio for columns and beams in Fig. 8(a) and set n = K c /K}). The deflection of the columns in Fig. 8(a) is similar to that of member AD in Fig. 27 for which/ = +1; therefore, K c {2 + /) = 3K C . The beams in Fig. 8(a) deflect as member AC in Fig. 27 for which / = —1, and Kb(2 + /) = Kb. Insert SK C and Kb in the equation in Fig. 27 which gives the relationship between 6 and U in Fig. 8(a) : 2EQ = (27) 2(SK C ) + 2(2Q,) The end moment, M, induced by joint rotations as in member AD in Fig. 27 is M = 3K c (2Ee), which combined with equation (27) gives M = Kr „ W ' rjr = n Zn n XU (28) C 6K C + 2Kb 6n + 2 At exterior columns, with loading arranged as in Fig. 8(c) to give approx- imately maximum column moment, the value of K c (2 + /) equals 3A' C as for interior columns; but the shape of the deflected beam in the end span is somewhere between the shapes of AB(f = 0) and AC (J = — 1) in Fig. 27. Adopting/ = -0.5 gives iQ>(2 +/) = 1.5Kb, and equations (27) and (28) become modified as follows : 2ge - 2(3* C ) +!(!.«») ^ and M = — ^— X U (30) 4n + 1 APPENDIX B: DERIVATION OF FORMULAS- WIND PRESSURE 29. Assumptions and Distribution of Wind Pressure Let A f B, C, D and F in Fig. 30 be joints in a bent which is deformed by bending due to wind pressure. While investigating the conditions around joint A, make the following assumptions:* (a) joints F, A and C lie on a straight line (b) joints B, A and D lie on a straight line (c) the same angle change, B A , exists at F, A and C, 0^ being measured from a horizontal line (d) the same angle change, G A , exists at B } A and D, A being measured from vertical lines. *Suggested by Professors Wilson and Maney on page 25 in Reference 21 . 53 The assumptions, incorporated in Fig. 30, place the point of contraflexure at the midpoints of all members. They also provide for a method of distri- bution of wind pressure as derived in the paragraphs which follow. The part of the bent shown diagrammatically in Fig. 30 will deflect under wind pressure, being the angle change at the joints and R an angle representing translation of the joints. 3 Moments induced by joint translation Fig. 30 The sum of the moments induced in the members at A by the joint translation equals zero because A is in equilibrium and no forces are applied between the joints. The induced moments, expressed by equation (21), are written below, the signs being in accordance with convention (1) in Section 27. Fig. 30(a) shows that the moments in the beams tend to rotate A in one direction and the moments in the columns tend to rotate A in the opposite direction. It is, therefore, the sum of the four moments within the parentheses that equals zero. M AC M AF M AB M AD +2BK AC X[2e A + { + eA)} -2EK AF X[26 A + (+e A )] +2EK AB X[-3(R-e A )] -2EK AD X[~3(R-e A )] + (6EK AC XQ A ) -(6EK AF XS A ) + (6EK AB XQ A - - (6EK AD X Q A ■ 6EK AB X R) 6EK AD XR) 2M AX = Q = 6Ee A X2(K AX )-GERX(K AB +K AD ). 54 The equation below the summation line reduces to = e A X (2K AX ) - R(K AB + K AD ), from which 6. = R X K AB + K AD ZK (31) AX Combine the equation p _ a = 7? v (E—**l ~ ^ AB - Z ^p) = /? v — ^ ~^~ ^ AF " Tr 4i 2K A x with M AB = -$EK AB (R - 9 A ), which gives ^ac + Kaf M AB = -§ERXK AB -^-^^ (32) &&-AX If the shear in column AB is denoted as V A , then V A =\XM AB = -^? X K AB (^+J^), . . . (33) in which the numerator within the parenthesis is the sum of the 7C-values for the beams at A and the denominator is the sum of the i£-values for all members at A. The value of R, the only unknown in equations (31), (32) and (33), need not be determined when R/h is constant for all columns in a story. In this case, the relative values, v A , of the shears, V A , taken by each column become equal to -rr Kac + Kaf /q,f\ v a . = Kab —^ (<*4) In addition, the sum of all column shears in a story equals the total wind shear. From this requirement together with equation (34) , shears and sub- sequently moments may be calculated in the columns. Shears and moments may then be computed in the beams by using the assumption of contra- flexure at mid span. Let the dotted line extending downward past D in Fig. 30 indicate that AD' is a basement story column, with height h' and assumed fixed at the footing, D'. The regular distribution of wind pressure may be applied in this case also, except that for columns as AD' } the effective story height, h, should be taken somewhat smaller than the actual story height h', say, h = % X h'. ♦Equation (34) gives shear in column AB above joint A. For the column below, AD, substitute K AD for K AB in equation (34). r ,5 BIBLIOGRAPHY Vertical Load Analysis 1. "Analysis of Statically Indeterminate Structures by the Slope-Deflection Method/' W. M. Wilson, F. E. Richart and Camillo Weiss, Bulletin 108, Univ. of 111., Eng. Exp. Sta., 1918. 2. "Reinforced Concrete Design," Oscar Faber, published by Arnold, London, Vol. 2, 1924. 3. "Statically Indeterminate Stresses," Parcel and Maney, published by John Wiley & Sons, 1926. 4. "Moments in Restrained and Continuous Beams by the Method of Conjugate Points," L. H. Nishkian and D. B. Steinman, Trans. AJS.C.E., 1927, Vol. 90, pp. 1-206 (inch discussion). 5. "Continuity as a Factor in Reinforced Concrete Design," Hardy Cross, Proc. A.CJ. 9 1929, pp. 669-711. 6. "Simplified Rigid Frame Design," Hardy Cross, Proc. A.C.I., Vol 26, 1930, pp. 170-183. 7. "Standards of Design for Concrete," U. S. Navy Dept., Bureau of Yards and Docks, U. S. Government Printing Office, Washington, 1930. 8. "The Column Analogy," Hardy Cross, Bulletin 215, Univ. of 111., Eng. Exp. Sta., 1930. 9. "Vertical-Load Analysis of Rigid Building Frames Made Practicable," John E. Goldberg, Engineering News-Record, November 12, 1931, pp. 770-772. 10. "Analysis of Continuous Frames by Distributing Fixed End Moments/' Hardy Cross, Tram. A.S.C.E., 1932, Vol. 96, pp. 1-156 (inch discussion). 11. "Continuous Frames of Reinforced Concrete," Cross and Morgan, published by John Wilev & Sons, 1932. 12. "The Modified Slope-Deflection Equations," L. T. Evans, Proc. A.C.L, 1932, pp. 109-130. 13. "Design of Continuous Frames Having Variable Moment of Inertia,' ' Thor Germundsson, Civil Engineering, October, 1932, pp. 647-648. 14. "The Design of Steel Mill Buildings and the Calculation of Stresses in Framed Structures," Milo S. Ketchum, published by McGraw-Hill, 1932. 15. " Continuous-Beam Analysis by Direct Distribution," J. C. Schulze and W. A. Rose, Engineering News-Record, December 20, 1934, pp. 782-785. 16. "Analysis of Continuous Structures by Traversing the Elastic Curves," Ralph W. Stewart, Proc. A.S.C.E., October, 1934, pp. 1125-1134. 17. "Handbook of Rigid Frame Analysis," L. T. Evans, printed by Edwards Brothers, Inc., Ann Arbor, Mich., 1934. 18. "Analysis of Continuous Frames by the Method of Restraining Stiffnesses," by E. B. Russell, published by Ellison and Russell, San Francisco, Calif., 1934. 19. "A Direct Method of Moment Distribution," T. Y. Lin, Proc. A.S.C.E., December, 1934, pp. 1451-1461. (See discussion by L. E. Grinter, March, 1935, p. 425.) 20. "Rigid-Frame Design in the Bureau of Yards and Docks," G. A. Hunt, Engineering News-Record, June 27, 1935, pp. 915-917. 56 BIBLIOGRAPHY Wind Stress Analysis 21. "Wind Stresses in the Steel Frames of Office Buildings," W. M. Wilson and G. A. Maney, Bulletin 80, Univ. of III., Eng. Exp. Sta., 1915. 22. "Wind Stresses in the Steel Frames of Office Buildings," Albert Smith and W. M. Wilson, Jl. Western Soc. of Eng., Apr., 1915, pp. 341-390. 23. "Column Deflection Method for Designing Lateral Bracing," John C. Van der Mey and Felix H. Spitzer, Engineering News-Record, January 19, 1928, pp. 106-108. 24. "Design of Tall Building Frames to Resist Wind," Morris and Rose, Bulletin 48, Ohio State Univ., 1929. 25. "Wind Stresses in Tall Buildings," R. Fleming, Engineering and Contracting, June, 1929, pp. 250-252. 26. "Wind Stresses in Buildings," R. Fleming, published by John Wiley & Sons, 1930. 27. "Wind on Tall Buildings," E. E. Seelye, Engineering News-Record, March 20, 1930. pp. 481-483. 28. "Wind Bracing: The Importance of Rigidity in High Towers," H. V. Spurr, published by McGraw-Hill, 1930. 29. Expert Testimony on Wind Design for Tall Buildings: I. "A Review of Current Theories," D. C. Coyle, Engineering News-Record, June 4, 1931, pp. 932-934. II. "Assumed Loads and Fiber Stresses," Albert Smith, Engineering News- Record, June 11, 1931, pp. 971-974. III. "Chord Deflections Control Web System Design," Henry V. Spurr, Engineer- ing News-Record, June 18, 1931, pp. 1012-1014. IV. "An Analysis of Some Basic Assumptions," C. T. Morris, Engineering News- Record, June 25, 1931, pp. 1050-1051. 30. "Wind Stresses on Portal Frames," W. S. Gray, Concrete and Consir. Eng., 1932, pp. 623-631. 31. "Deflections and Vibrations in High Buildings," L. J. Mensch, Proc. A.C.J., Vol. 28, 1932, pp. 387-404. 32. "Wind-Stress Analysis and Moment Distribution," R. Fleming, Engineering News- Record, February 9, 1933, pp. 194-195. 33. "Wind-Bracing Problems," A. Smith, Jl. Western Soc. of Eng., February, 1933, pp. 1-18. 34. "Wind Stresses by Slope Deflection and Converging Approximations," J. E. Goldberg, Trans. A.S.C.E., 1934, pp. 962-985. 35. "Wind Stress Analysis Simplified," L. E. Grinter, Trans. A.S.C.E., 1934, pp. 610-669. Discussion closed in Proc. A.S.C.E., January, 1934, pp. 93-102. 36. "Basis of Design for Hurricane Exposure," Albert Smith, Proc. A.C.I. , Vol. 27, 1931, pp. 903-924. 37. "Structural Frameworks," Thomas F. Hickerson, The University of North Carolina Press, 1934. 57 NOTES PORTLAND CEMENT ASSOCIATION A National Organization to Improve and Extend the Uses of Concrete 3 3 WEST GRAND AVENUE » CHICAGO PRINTED IN U. S. A* T-20-UM- 12-86 , [